25 Points! Please answer asap! Carly stated “All pairs of rectangles are dilations”. Which pair of rectangles would prove that Carly’s statement is incorrect? (Images below)

25 Points! Please Answer Asap! Carly Stated All Pairs Of Rectangles Are Dilations. Which Pair Of Rectangles
25 Points! Please Answer Asap! Carly Stated All Pairs Of Rectangles Are Dilations. Which Pair Of Rectangles

Answers

Answer 1

Answer:

C

Step-by-step explanation:

A. First two rectangles are dilations because

[tex]\dfrac{2}{4}=\dfrac{4}{8}=0.5[/tex]

B. Second two rectangles are dilations because

[tex]\dfrac{2}{4}=\dfrac{3}{6}=0.5[/tex]

C. Third two rectangles are not dilations because

[tex]\dfrac{3}{4}\neq \dfrac{2}{3}[/tex]

D. Fourth two rectangles are dilations because

[tex]\dfrac{3}{4}=\dfrac{1.5}{2}=0.75[/tex]

Answer 2

Answer:

c please correct me if im wrong

Step-by-step explanation:


Related Questions

find integers x and y such that 115x + 30y = gcd(431, 29)

Answers

Answer:

There are none.

Step-by-step explanation:

gcd(431,29)=1  

The reasons I came to this conclusion is because 29 is prime.

29 is not a factor of 431 so we are done.

So now we want to find (x,y) such that 115x+30y=1.

I'm going to use Euclidean's Algorithm.

115=30(3)+25

30=25(1)+5

25=5(5)

So we know we are done when we get the remainder is 0 and I like to look at the line before the remainder 0 line to see the greatest common divisor or 115 and 30 is 5.

So 115x+30y=5 has integer solutions (x,y) where d=5 is the smallest possible positive such that 115x+30y=d will have integer solutions (x,y).

So since 1 is smaller than 5 and we are trying to solve 115x+30y=1 for integer solutions (x,y), there there is none.

Furthermore,  115x+30y=1 can be written as 5(23x+6y)=1 and we know that 5 is not a factor of 1.

The random variables X and Y have the joint PMF pX,Y(x,y)={c⋅(x+y)2,0,if x∈{1,2,4} and y∈{1,3},otherwise. All answers in this problem should be numerical. Find the value of the constant c .

Answers

In order for [tex]p_{X,Y}(x,y)[/tex] to be a valid PMF, its integral over the distribution's support must be equal to 1.

[tex]p_{X,Y}(x,y)=\begin{cases}\dfrac{c(x+y)}2&\text{for }x\in\{1,2,4\}\text{ and }y\in\{1,3\}\\\\0&\text{otherwise}\end{cases}[/tex]

There are 3*2 = 6 possible outcomes for this distribution, so that

[tex]\displaystyle\sum_{x,y}p_{X,Y}(x,y)=\sum_{x\in\{1,2,4\}}\sum_{y\in\{1,3\}}\frac{c(x+y)}2=1[/tex]

[tex]1=\displaystyle\frac c2\sum_{x\in\{1,2,4\}}((x+1)+(x+3))=\frac c2\sum_{x\in\{1,2,4\}(2x+4)[/tex]

[tex]1=\displaystyle\frac c2((2+4)+(4+4)+(8+4))[/tex]

[tex]1=13c\implies\boxed{c=\dfrac1{13}}[/tex]

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. (If an answer does not exist, enter DNE.) x = t2 − t + 9, y = t3 − 3t

Answers

Answer:

The horizontal tangents occur at: (9,-2) and (11,2)

The vertical tangent occurs at (8.75,-1.375)

See attachment

Step-by-step explanation:

The given parametric equations are:

[tex]x=t^2-t+9[/tex] and [tex]y=t^3-3t[/tex]

The slope function is given by:

[tex]\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]

[tex]\frac{dy}{dx}=\frac{3t^2-3}{2t-1}[/tex]

The tangent is vertical when [tex]\frac{dx}{dt}=0[/tex]

[tex]\implies 2t-1=0[/tex]

[tex]t=\frac{1}{2}[/tex]

When [tex]t=\frac{1}{2}[/tex], [tex]x=(\frac{1}{2})^2-\frac{1}{2}+9=8.75[/tex], [tex]y=0.5^3-3(0.5)=-1.375[/tex]

The vertical tangent occurs at (8.75,-1.375)

The tangent is horizontal when [tex]\frac{dy}{dt}=0[/tex]

[tex]3t^2-3=0[/tex]

[tex]\implies t=\pm1[/tex]

When t=1, [tex]x=(1)^2-1+9=9[/tex], [tex]y=1^3-3(1)=-2[/tex]

When t=-1, [tex]x=(-1)^2+1+9=11[/tex], [tex]y=(-1)^3-3(-1)=2[/tex]

The horizontal tangents occur at: (9,-2) and (11,2)

Find the y -intercept and the slope of the line.
Write your answers in simplest form

5x - 2y = 2

Answers

Answer:

The slope is 5/2 and the y intercept is -1

Step-by-step explanation:

To find the slope and the y intercept, we will write the equation in slope intercept form, y =mx+b where m is the slope and b is the y intercept

5x -2y =2

Add 2y to each side

5x-2y+2y =2 +2y

5x = 2+2y

Subtract 2 from each side

5x-2 = 2y+2-2

5x-2 =2y

Divide each side by 2

5x/2 -2/2 = 2y/2

5/2x -1 = y

y = 5/2x -1

The slope is 5/2 and the y intercept is -1

Comparing yourself to a triangle with 3 possible occupation/career choices, could this be compared to "congruent" rather than "equal"? Explain your thinking...

Answers

Answer:

Yes

Step-by-step explanation:

If I am understanding the question correctly, then I would have to say it is compared to "congruent".

Congruent: two triangles are compared and have the same sides and angles

Equilateral: all 3 sides of the triangle are the same.

Basing my opinion on the definitions above. I believe this situation could be compared to "congruent" because as mentioned in the question we are comparing ourselves to a triangle that has 3 DIFFERENT career choices.

If this was compared to "Equal" there would be 1 career choice repeated 3 times.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

22 Which of the following polynomials has a factor of x-1? A) p(x)=x^3 +x^2 -2x+1 B) q(x)=2x^3-x^2 +x-1 (Crx)= 3x^3-x-2 D) S(x)=-3x^3+ 3x +1

Answers

Answer:

The correct option is C.

Step-by-step explanation:

If (x-c) is a factor of a polynomial f(x), then f(c)=0.

It is given that (x-1) is a factor of the polynomial. It means the value of the function at x=1 is 0.

In option A,

The given function is

[tex]p(x)=x^3+x^2-2x+1[/tex]

Substitute x=1 in the given function.

[tex]p(1)=(1)^3+(1)^2-2(1)+1=1+1-2+1=1[/tex]

Since p(1)≠0, therefore option A is incorrect.

In option B,

The given function is

[tex]q(x)=2x^3-x^2+x-1[/tex]

Substitute x=1 in the given function.

[tex]q(1)=2(1)^3-(1)^2+(1)-1=2-1+1-1=1[/tex]

Since q(1)≠0, therefore option B is incorrect.

In option C,

The given function is

[tex]r(x)=3x^3-x-2[/tex]

Substitute x=1 in the given function.

[tex]r(1)=3(1)^3-(1)-2=3-1-2=0[/tex]

Since r(1)=0, therefore option C is correct.

In option D,

The given function is

[tex]s(x)=-3x^3+3x+1[/tex]

Substitute x=1 in the given function.

[tex]s(1)=-3(1)^3+3(1)+1=-3+3+1=1[/tex]

Since s(1)≠0, therefore option D is incorrect.

An investor can invest money with a particular bank and earn a stated interest rate of 15.40%; however, interest will be compounded quarterly. What are the nominal, periodic, and effective interest rates for this investment opportunity?

Answers

Answer:

Nominal rate = 15.40%, Periodic rate = 3.85% and effective interest rate = 16.31%

Step-by-step explanation:

An investor can invest money and earn a stated interest rate = 15.40%

We have to find the nominal, periodic and effective interest rates for this investment.

Nominal rate = Stated interest rate is also called annual percentage rate or nominal interest rate. Therefore nominal interest rate = 15.40%.

Periodic interest rate = Periodic interest rate is the annual interest rate divided by number of compounding periods.

Periodic rate = ( [tex]\frac{15.40}{4}[/tex] ) = 3.85%

Effective interest rate = We have to calculate Effective interest rate (EAR)  by the formula =  [tex]EAR=[(\frac{1+i}{n})^{n}-1][/tex]

where i = rate of interest and n = number of compounding periods

[tex]EAR=[(1+\frac{0.154}{4})^4]-1[/tex]

[tex]EAR=[(1+.0385)^{4}]-1[/tex]

[tex]EAR=[(1.0385)^{4}]-1[/tex]

EAR = 1.1631 - 1

EAR = 0.1631 = 16.31%

Therefore, Nominal rate = 15.40%, Periodic rate = 3.85% and effective interest rate = 16.31%

A rumor spreads among a group of 400 people. The number of people, N(t), who have heard the rumor by the time t in hours since the rumor started to spread can be approximated by a function of the form
N(t) = 400 / (1+399e^(-0.4t))
Approximately how long will it take until half the people have heard the rumor?

Answers

Answer:

  about 15 hours

Step-by-step explanation:

You want to find t such that N(t)=200. Fill in the equation with that information and solve for t.

  200 = 400/(1 +399e^(-0.4t))

  1 +399e^(-0.4t) = 400/200 = 2 . . . . . multiply by (1+399e^(-0.4t))/200

  399e^(-0.4t) = 1 . . . . . . . . . . . . . . . . . . subtract 1

  e^(-0.4t) = 1/399 . . . . . . . . . . . . . . . . . .divide by 399

  -0.4t = ln(1/399) . . . . . . . take the natural log

  t = ln(399)/0.4 ≈ 14.972 . . . . . . . divide by -0.4, simplify

Rounded to tenths, it will take 15.0 hours for half the people to have heard the rumor.

Final answer:

The time it will take for half the group, or 200 people, to have heard the rumor based on the given exponential decay function is approximately 1.72 hours.

Explanation:

The problem presented is a mathematical question involving a model of exponential decay. In this context, we want to know when half of the people, or 200 people, will have heard the rumor. To find out, we substitute N(t) with 200 in the equation given:

200 = 400 / (1+399e^(-0.4t))

After simplifying and solving for t, t comes out to be approximately 1.72 hours. So, it takes approximately 1.72 hours for half the people to have heard the rumor.

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Prove (a) cosh2(x) − sinh2(x) = 1 and (b) 1 − tanh 2(x) = sech 2(x). SOLUTION (a) cosh2(x) − sinh2(x) = ex + e−x 2 2 − 2 = e2x + 2 + e−2x 4 − = 4 = . (b) We start with the identity proved in part (a): cosh2(x) − sinh2(x) = 1. If we divide both sides by cosh2(x), we get 1 − sinh2(x) cosh2(x) = 1 or 1 − tanh 2(x) = .

Answers

Answer:

See explanation.

Step-by-step explanation:

The hyperbolic sine and cosine functions are defined as follows:

[tex] \sinh(x) = \frac{ {e}^{x} - {e}^{ - x} }{2} [/tex]

[tex]\cosh(x) = \frac{ {e}^{x} + {e}^{ - x} }{2} [/tex]

We want to show that:

[tex]\cosh^{2} (x) - \sinh^{2} (x) = 1[/tex]

We use the definition by substituting the expressions into the left hand side and simplify to obtain the RHS.

[tex] \cosh^{2} (x) - \sinh^{2} (x) = {( \frac{ {e}^{x} + {e}^{ - x} }{2} )}^{2} + {( \frac{ {e}^{x} - {e}^{ - x} }{2} )}^{2} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} +2 {e}^{x} {e}^{ - x} + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 {e}^{x} {e}^{ - x} + {e}^{ - 2x} }{4} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} +2 + {e}^{ - 2x} }{4} - \frac{ {e}^{2x} - 2 + {e}^{ - 2x} }{4} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ {e}^{2x} + {e}^{ - 2x} - {e}^{2x} + {e}^{ - 2x }+ 2 +2 }{4} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x) = \frac{ 4 }{4} [/tex]

[tex]\cosh^{2} (x) - \sinh^{2} (x)=1[/tex]

b)

If we start with the identity in a) and we divide both sides by cosh²x we get:

[tex] \frac{\cosh^{2} (x) }{\cosh^{2} (x) } -\frac{\sinh^{2} (x) }{\cosh^{2} (x) } =\frac{1}{\cosh^{2} (x) } [/tex]

This simplifies to:

[tex]1 - \tanh ^{2} (x) = \sec \: h ^{2} (x) [/tex]

There are 40 students in a course: 15 freshmen, 8 sophomores, 12 juniors, and 5 seniors. One student is selected at random to attend a workshop and report back to the class. What is the probability that the chosen student is a junior

Answers

Answer:

The probability that the chosen student is a junior is [tex]\frac{3}{10}[/tex]

Step-by-step explanation:

Given,

The total number of students = 40,

In which, freshmen = 15,

Sophomores = 8,

Juniors = 12,

And, seniors = 5,

Thus, the probability that junior student when a student is chosen

[tex]=\frac{\text{Total possible arrangement of 1 junior student }}{\text{Total possible arrangement of 1 student}}[/tex]

[tex]=\frac{^{12}C_1}{^{40}C_1}[/tex]

[tex]=\frac{12}{40}[/tex]

[tex]=\frac{3}{10}[/tex]

Final answer:

The probability of selecting a junior student is 0.3 or 30%.

Explanation:

To find the probability of selecting a junior student, we need to divide the number of junior students by the total number of students. There are 12 junior students out of a total of 40 students, so the probability can be calculated as:

Probability = Number of junior students / Total number of students

Probability = 12 / 40

Probability = 0.3

Therefore, the probability of selecting a junior student is 0.3 or 30%.

Learn more about probability here:

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Insert parentheses to make the following sentence true. 4 • 5 + 16 - 2 = 82

Answers

Answer:

[tex]4 \cdot(5 + 16) - 2 = 82[/tex]

Step-by-step explanation:

The given equation is

[tex]4 \cdot5 + 16 - 2 = 82[/tex]

We want to insert parenthesis, so that this equation will be true.

We insert parenthesis to obtain:

[tex]4 \cdot(5 + 16) - 2 [/tex]

We now use PEDMAS to simplify.

[tex]4 \cdot(21) - 2 =82[/tex]

[tex]84 - 2 = 82[/tex]

[tex]82 = 82[/tex]

The Earth is 4.6 billion years old, but human civilization (the building of cities) began only about 10,000 years ago. If you represent the history of Earth with a line one mile long (63,360 inches), how long must the line be to represent the history of human civilization?

Answers

Answer:

0.13774 inches ( approx ).

Step-by-step explanation:

Given,

The age of earth = 4.6 billion years = 4600000000 years,

The represented age of earth = 63,360 inches,

That is, scale factor in representation of age

[tex]=\frac{\text{Represented age}}{\text{Actual age}}[/tex]

[tex]=\frac{ 63,360}{4600000000}[/tex]

Now, the age of human civilization = 10000 years,

Thus, the represented age of human civilization = actual age × scale factor

[tex]=10000\times \frac{ 63,360}{4600000000}[/tex]

[tex]=\frac{633600000}{4600000000}[/tex]

[tex]=0.137739130435[/tex]

[tex]\approx 0.13774\text{ inches}[/tex]

(1.)Find the slope of the line that passes through the given pair of points. (If an answer is undefined, enter UNDEFINED.) (?a + 3, b ? 3) and (a + 3, ?b) *******(2.)If the line passing through the points (a, 1) and (6, 5) is parallel to the line passing through the points (2, 7) and (a + 2, 1), what is the value of a?

Answers

Answer:

1. The slope of the line is [tex]m=\frac{-2b+3}{2a}[/tex].

2. The value of a is 18.

Step-by-step explanation:

If a line passes through two points, then the slope of the line is

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

(1)

It is given that the line passes through the points (-a + 3, b - 3) and (a + 3, -b). So, the slope of the line is

[tex]m=\frac{-b-(b-3)}{a+3-(-a+3)}[/tex]

[tex]m=\frac{-b-b+3}{a+3+a-3)}[/tex]

[tex]m=\frac{-2b+3}{2a}[/tex]

The slope of the line is [tex]m=\frac{-2b+3}{2a}[/tex].

(2)

If the line passing through the points (a, 1) and (6, 5), then the slope of the line is

[tex]m_1=\frac{5-1}{6-a}=\frac{4}{6-a}[/tex]

If the line passing through the points (2, 7) and (a + 2, 1), then the slope of the line is

[tex]m_2=\frac{1-7}{a+2-2}=\frac{-6}{a}[/tex]

The slopes of two parallel lines are same.

[tex]m_1=m_2[/tex]

[tex]\frac{4}{6-a}=\frac{-6}{a}[/tex]

On cross multiplication we get

[tex]4a=-6(6-a)[/tex]

[tex]4a=-36+6a[/tex]

[tex]4a-6a=-36[/tex]

[tex]-2a=-36[/tex]

Divide both sides by -2.

[tex]a=18[/tex]

Therefore the value of a is 18.

1) For the following problem: Let f={(-2,3),(-1,1),(0,0),(1,-1), (2,-3)} and let g-{-3,1),(-1,-2), (0, 2),(2, 2),(3,1)}.Find the following a) f(1) and g-1) b) (gof (1) c) (gofof)(-1) d) (fog)(3)

Answers

Answer:

1. f(1)=-1 and g(-1)=-2.

2. (gof)(1)=-2.

3. (gofof)(-1)=-2

4. (fog)(3)=-1

Step-by-step explanation:

The given functions are defined as

f={(-2,3),(-1,1),(0,0),(1,-1), (2,-3)}

g={(-3,1),(-1,-2), (0, 2),(2, 2),(3,1)}

1.

The value of function f at x=1 is -1, So, f(1)=-1.

The value of function g at x=-1 is -2, So, g(-1)=-2.

Therefore the value of f(1) is -1 and the value of g(-1) is -2.

2.

[tex](g\circ f)(1)=g(f(1))[/tex]                       [tex][\because (g\circ f)(x)=g(f(x))][/tex]  

[tex](g\circ f)(1)=g(-1)[/tex]                        [tex][\because f(1)=-1][/tex]  

[tex](g\circ f)(1)=-2[/tex]                           [tex][\because g(-1)=-2][/tex]  

Therefore the value of (gof)(1) is -2.

3.

[tex](g\circ f\circ f)(-1)=(g\circ f)(f(-1))[/tex]                       [tex][\because (g\circ f)(x)=g(f(x))][/tex]  

[tex](g\circ f\circ f)(-1)=(g\circ f)(1)[/tex]                        [tex][\because f(-1)=1][/tex]  

[tex](g\circ f\circ f)(-1)=-2[/tex]                           [tex][\because \text{From part 2}, (g\circ f)(1)=-2][/tex]  

Therefore the value of (gofof)(1) is -2.

4.

[tex](f\circ g)(3)=f(g(3))[/tex]                       [tex][\because (f\circ g)(x)=f(g(x))][/tex]  

[tex](f\circ g)(3)=f(1)[/tex]                        [tex][\because g(3)=1][/tex]  

[tex](f\circ g)(3)=-1[/tex]                           [tex][\because f(1)=-1][/tex]  

Therefore the value of (fog)(3) is -1.

Let A, B, and C be arbitrary sets within a universal set, U. For each of the following statements, either prove that the statement is always true or show a counterexample to prove it is not always true. When giving a counterexample, you should define the three sets explicitly and say what the left-hand and right-hand sides of the equation are for those sets, to make it clear that they are not equal.

(A \ B) × C = (A × C) \ (B × C)

Answers

Answer with Step-by-step explanation:

Let A, B and C are arbitrary sets within a universal set U.

We  have to prove that [tex]( A/B)\times C=(A\times C)/(B\times C)[/tex] is always true.

Let [tex](x,y)\in (A/B)\times C[/tex]

Then [tex] x\in(A/B) [/tex] and [tex] y\in C[/tex]

Therefore, [tex] x\in A[/tex] and [tex] x\notin B[/tex]

Then, (x,y) belongs to [tex] A\times C[/tex]

and (x,y) does not belongs to [tex] B\times C[/tex]

Hence,[tex] (x,y)\in(A\times C)/(B\times C)[/tex]

Conversely ,Let (x ,y)belongs to [tex] (A\times C)/(B\times C)[/tex]

Then [tex] (x,y)\in (A\times C)[/tex] and [tex] (x,y)\notin (B\times C)[/tex]

Therefore,[tex] x\in A,y\in C[/tex] and [tex] x\notin B,y\in C[/tex]

[tex] x\in(A/B)[/tex] and [tex]y\in C[/tex]

Hence, [tex] (x,y)\in(A/B)\times C[/tex]

Therefore,[tex] (A/B)\times C=(A\times C)/(B\times C)[/tex] is always true.

Hence, proved.

Pulam, Inc. prepared the following master budget items for July:
Production and sales 36,000 units
Variable manufacturing costs:
Direct materials $ 36,000
Direct labor $ 72,000
Variable manufacturing overhead $ 72,000
Fixed manufacturing costs $ 180,000
Total manufacturing costs $ 360,000
During July, Pulam actually sold 42,000 units. Prepare a flexible budget for Pulam based on actual sales.(Do not round your intermediate calculations).

Production and sales units
Variable manufacturing costs:
Direct materials $
Direct labor $
Variable manufacturing overhead $
Fixed manufacturing costs $
Total manufacturing costs $

Answers

Answer:

During July, Pulam actually sold 42,000 units.

Variable Manufacturing Costs for 42,000 units  are-

1. Direct Material : [tex](36000/36000)\times42000 =42000[/tex] dollars

2. Direct Labor:  [tex](72000/36000)\times42000 =84000[/tex] dollars

3. Variable Manufacturing Overhead : [tex](72000/36000)\times42000 =84000[/tex] dollars

4. Fixed manufacturing costs : $ 180000

5. Total manufacturing costs : [tex]42000+84000+84000+180000=390000[/tex] dollars

Find an equation for the tangent to the curve at P and the horizontal tangent to the curve at Q. y = 5 + cot x - 2 csc x 0 1 2 3 0 2 4 x y Upper QUpper P left parenthesis StartFraction pi Over 2 EndFraction comma 3 right parenthesis

Answers

Answer with explanation:

The given function in x and y is,

  y= 5 +cot x-2 Cosec x

To find the equation of tangent, we will differentiate the function with respect to x

[tex]y'= -\csc^2 x+2 \csc x\times \cot x[/tex]

Slope of tangent at (π/2,3)

 [tex]y'_{(\frac{\pi}{2},3)}= -\csc^2\frac{\pi}{2} +2 \csc \frac{\pi}{2}\times \cot \frac{\pi}{2}\\\\=-1+2\times 1 \times 0\\\\= -1[/tex]

Equation of tangent passing through (π/2,3) can be obtained by

[tex]\rightarrow \frac{y-y_{1}}{x-x_{1}}=m(\text{Slope})\\\\ \rightarrow \frac{y-3}{x-\frac{\pi}{2}}=-1\\\\\rightarrow 3-y=x-\frac{\pi}{2}\\\\\rightarrow x+y-3-\frac{\pi}{2}=0[/tex]

⇒There will be no Horizontal tangent from the point (π/2,3).

Final answer:

The equation of the tangent to the curve y = 5 + cot x - 2 csc x at a given point can be found using the derivative, but for point P (π/2, 3), the slope is undefined because cot x and csc x are undefined at π/2. A horizontal tangent occurs where the derivative is zero, and point Q must be determined by solving y' = 0.

Explanation:

To find the equation of the tangent at point P, we first need to compute the derivative of the given function y = 5 + cot x - 2 csc x to find the slope of the tangent at any point on the curve. For the given curve y = 5 + cot x - 2 csc x, the derivative y' will give us the slope of the tangent at any point x. The slope of the tangent at point P, where P is (π/2, 3), can be found by evaluating the derivative at x = π/2.

However, since cot x and csc x are undefined at x = π/2, this curve will not have a well-defined slope and therefore not a conventional tangent line at x= π/2. On the other hand, a horizontal tangent will occur at points where the derivative y' is equal to zero. To find point Q, we must solve y' = 0 for x, and then use the x-value to find the corresponding y-coordinate on the curve.

Which established actor stared in The Godfather as Don Vito? a. Marlon Brando b. John Wayne c. Henry Fonda d. None of the above

Answers

Answer:

a) Marlon Brando

Step-by-step explanation:

Don Vito Corleone was played by Marlon Brando in the film Godfather released in the year 1972 directed by Francis Ford Coppola based on the 1969 book by Mario Puzo. Paramount studios was against the casting of Marlon Brando but the director insisted on giving him a screen test. The screen test convinced the studio to cast him in the role.

Final answer:

Marlon Brando starred as Don Vito Corleone in The Godfather, earning an Academy Award for his performance.

Explanation:

The established actor who starred in The Godfather as Don Vito Corleone is Marlon Brando. Brando's performance in this iconic film is widely recognized as one of his greatest roles. Released in 1972, The Godfather is a crime film directed by Francis Ford Coppola and based on the novel of the same name by Mario Puzo. Marlon Brando's portrayal of the mafia patriarch earned him an Academy Award for Best Actor, which further solidified his legacy as one of Hollywood's legendary actors. The other actors listed, John Wayne and Henry Fonda, were not part of the cast of The Godfather.

The Vertex of a parabola is at (8-1), and it's why intercept is negative 17, which function represents the parabola

Answers

Answer:

The function is equal to [tex]y=-(1/4)(x-8)^{2}-1[/tex]

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

[tex]y=a(x-h)^{2}+k[/tex]

where

a is a coefficient

(h,k) is the vertex

In this problem we have

(h,k)=(8,-1)

substitute

[tex]y=a(x-8)^{2}-1[/tex]

Find the value of a

Remember that we have the y-intercept

The y-intercept is the point (0,-17)

substitute

x=0,y=-17

[tex]-17=a(0-8)^{2}-1[/tex]

[tex]-17=64a-1[/tex]

[tex]64a=-17+1[/tex]

[tex]64a=-16[/tex]

[tex]a=-16/64[/tex]

[tex]a=-1/4[/tex]

therefore

The function is equal to

[tex]y=-(1/4)(x-8)^{2}-1[/tex]

see the attached figure to better understand the problem

Find the equation of the line that passes through the points (12,21) and (46,33). V.35r-162

Answers

Answer:

the answer is Y=(6/17)X + (285/17)

Step-by-step explanation:

1. Identify the X and Y coordinate of each point.            

for example, the first point could be (12,21), therefore X1 = 12 and Y1 = 21 and the second point is then (46,33), therefore X2 = 46 and Y2 = 33

2. to find the equation of the line it is necessary to find the slope.

knowing that the equiation of the slope is M = Y2-Y1/X2-X1 we replace

  [tex]m=\frac{33-21}{46-12}=\frac{12}{34}[/tex] and then we simplify the result to [tex]\frac{12}{34}[/tex]

3. Using the Point-slope equation which is  Y-Y1=M(X-X1) we replace

 [tex]Y-33=\frac{6}{17}(X-46)[/tex]

[tex]Y=\frac{6}{17}X-\frac{276}{17}+33[/tex]

[tex]Y=\frac{6}{17}X+\frac{285}{17}[/tex]

Express the confidence interval left parenthesis 0.023 comma 0.085 right parenthesis(0.023,0.085) in the form of ModifyingAbove p with caret minus Upper E less than p less than ModifyingAbove p with caret plus Upper Ep−E

Answers

Final answer:

To express the confidence interval (0.023, 0.085) in the requested form, calculate the midpoint (p') as 0.054 and the margin of error (EBP) as 0.031. The confidence interval will thus be expressed as 0.054 - 0.031 < p < 0.054 + 0.031.

Explanation:

The question asks to express the confidence interval (0.023, 0.085) in the form of ModifyingAbove p with caret minus Upper E less than p less than ModifyingAbove p with caret plus Upper E. To do this, we first need to calculate the midpoint (p') and the margin of error (EBP) of the given interval.

The midpoint (p') is the average of the lower and upper bounds of the interval, calculated as (0.023 + 0.085) / 2 = 0.054. The margin of error (EBP) is the difference between p' and either boundary of the interval, calculated as 0.085 - 0.054 = 0.031 or 0.054 - 0.023 = 0.031. Therefore, the interval in the requested format is 0.054 - 0.031 < p < 0.054 + 0.031.

The confidence interval (0.023, 0.085) can be expressed in the form [tex]\( 0.054 - 0.031 < p < 0.054 + 0.031 \)[/tex].

To express the confidence interval (0.023, 0.085) in the form [tex]\( \hat{p} - E < p < \hat{p} + E \),[/tex] where [tex]\( \hat{p} \)[/tex] is the point estimate and ( E ) is the margin of error, we need to find the point estimate and the margin of error.

Step 1: Find the point estimate [tex]\( \hat{p} \)[/tex].

The point estimate is the midpoint of the confidence interval. We can calculate it using the formula:

[tex]\[ \hat{p} = \frac{\text{Lower bound} + \text{Upper bound}}{2} \][/tex]

Substitute the values:

[tex]\[ \hat{p} = \frac{0.023 + 0.085}{2} \][/tex]

[tex]\[ \hat{p} = \frac{0.108}{2} \][/tex]

[tex]\[ \hat{p} = 0.054 \][/tex]

So, the point estimate [tex]\( \hat{p} \)[/tex] is 0.054.

Step 2: Find the margin of error \( E \).

The margin of error is half the width of the confidence interval. We can calculate it using the formula:

[tex]\[ E = \frac{\text{Upper bound} - \text{Lower bound}}{2} \][/tex]

Substitute the values:

[tex]\[ E = \frac{0.085 - 0.023}{2} \][/tex]

[tex]\[ E = \frac{0.062}{2} \][/tex]

[tex]\[ E = 0.031 \][/tex]

So, the margin of error ( E ) is 0.031.

Step 3: Express the confidence interval in the required form.

Now, we can write the confidence interval in the form [tex]\( \hat{p} - E < p < \hat{p} + E \):[/tex]

[tex]\[ 0.054 - 0.031 < p < 0.054 + 0.031 \][/tex]

[tex]\[ 0.023 < p < 0.085 \][/tex]

Therefore, the confidence interval (0.023, 0.085) can be expressed in the form [tex]\( 0.054 - 0.031 < p < 0.054 + 0.031 \)[/tex].

solve y' -x^2y = 0 using power series and write the first four terms of the power series

Answers

We're looking for a solution of the form

[tex]y=\displaystyle\sum_{n\ge0}a_nx^n[/tex]

with derivative

[tex]y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n[/tex]

Note that [tex]x=0\implies y(0)=a_0[/tex].

Substituting into the ODE gives

[tex]\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n-\sum_{n\ge0}a_nx^{n+2}=0[/tex]

The first series starts with a constant term, while the second starts with a quadratic term, so we should pull out the first two terms of the first series and have it start at [tex]n=2[/tex], then shift the index on the second series to achieve the same effect, which allows us to condense the left side as

[tex]a_1+2a_2x+\displaystyle\sum_{n\ge2}\bigg((n+1)a_{n+1}-a_{n-2}\bigg)x^n=0[/tex]

so that the series solution's coefficients are given according to the recurrence

[tex]\begin{cases}a_0=a_0\\a_1=a_2=0\\(n+1)a_{n+1}-a_{n-2}=0&\text{for }n\ge2\end{cases}[/tex]

We can simplify the latter equation somewhat to get it in terms of [tex]a_n[/tex]:

[tex]a_n=\dfrac{a_{n-3}}n\text{ for }n\ge3[/tex]

This shows dependency between coefficients that are 3 indices apart, so we check 3 cases:

If [tex]n=3k+1[/tex], where [tex]k\ge0[/tex] is an integer, then

[tex]k=0\implies n=1\implies a_1=0[/tex]

[tex]k=1\implies n=4\implies a_4=\dfrac{a_1}4=0[/tex]

and so on for all such [tex]n[/tex], giving

[tex]a_{3k+1}=0[/tex]

If [tex]n=3k+2[/tex], then

[tex]k=0\implies n=2\implies a_2=0[/tex]

and we get the same conclusion as before,

[tex]a_{3k+2}=0[/tex]

If [tex]n=3k[/tex], then

[tex]k=0\implies n=0\impiles a_0=a_0[/tex]

[tex]k=1\implies n=3\implies a_3=\dfrac{a_0}3[/tex]

[tex]k=2\implies n=6\implies a_6=\dfrac{a_3}6=\dfrac{a_0}{3\cdot6}=\dfrac{a_0}{3^2(2\cdot1)}[/tex]

[tex]k=3\implies n=9\implies a_9=\dfrac{a_6}9=\dfrac{a_0}{3^3(3\cdot2\cdot1)}a_0[/tex]

and so on, with the general pattern

[tex]a_{3k}=\dfrac{a_0}{3^kk!}[/tex]

Then the series solution is

[tex]y=\displaystyle\sum_{k\ge0}\bigg(a_{3k}x^{3k}+a_{3k+1}x^{3k+1}+a_{3k+2}x^{3k+2}\bigg)[/tex]

[tex]y=\displaystyle a_0\sum_{k\ge0}\frac{x^{3k}}{3^kk!}[/tex]

[tex]y=\displaystyle a_0\sum_{k\ge0}\frac{\left(\frac{x^3}3\right)^k}{k!}[/tex]

whose first four terms are

[tex]\boxed{a_0\left(1+\dfrac{x^3}3+\dfrac{x^6}{18}+\dfrac{x^9}{162}\right)}[/tex]

When Bill makes a sandwich, he may choose from among 3 kinds of rolls, 4 varieties of meat, and 2 types of sliced cheese. If he chooses one roll, one meat, and one type of cheese, how many different kinds of sandwiches can he make?

Answers

Answer: Bill can make 24 different kinds of sandwiches

Step-by-step explanation:

Given : The number of kinds of rolls = 3

The number of varieties of meat = 4

The number of types of sliced cheese = 2

If he chooses one roll, one meat, and one type of cheese, then the number of different kinds of sandwiches he can make is given by :-

[tex]3\times2\times4=24[/tex]

Hence, Bill can make 24 different kinds of sandwiches.

Real estate ads suggest that 75 % of homes for sale have​ garages, 29 % have swimming​ pools, and 13 % have both features. What is the probability that a home for sale has ​a) a pool or a​ garage? ​b) neither a pool nor a​ garage? ​c) a pool but no​ garage?

Answers

Answer: a) 91%  b) 9% c) 16%

Step-by-step explanation:

Let A be the event that homes for sale have​ garages and B be the event that homes for sale have​ swimming​ pools.

Now, given :[tex]P(A)=75\%=0.75[/tex]

[tex]P(B)=29\%=0.29[/tex]

[tex]P(A\cap B)=13\%=0.13[/tex]

a) [tex]P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\\Rightarrow\ P(A\cup B)=0.75+0.29-0.13=0.91[/tex]

Hence, the probability that a home for sale has a pool or a​ garage is 91%.

b) The probability that a home for sale has ​neither a pool nor a​ garage is given by :-

[tex]1-P(A\cup B)=1-0.91=0.09[/tex]

Hence, the probability that a home for sale has ​neither a pool nor a​ garage is 9%.

c) The probability that a home for sale has a pool but no​ garage is given by :-

[tex]P(B)-P(A\cap B)=0.29-0.13=0.16[/tex]

Hence, the probability that a home for sale has a pool but no​ garage is 16%.

Final answer:

The probability a home for sale has a pool or garage is 91%, the probability of having neither feature is 9%, and the probability of having a pool but no garage is 16%.

Explanation:

The question involves solving problems related to probability and set theory. To find out the probability of events involving homes for sale with certain features, we can use the principle of inclusion and exclusion for probabilities.

Answering Part a)

The probability that a home has a pool or a garage is given by the formula P( A or B ) = P( A ) + P( B ) - P( A and B ). Thus, the probability is:

P( Pool or Garage ) = P( Pool ) + P( Garage ) - P( Pool and Garage )

= 0.29 + 0.75 - 0.13

= 0.91 or 91%

Answering Part b)

To find a home with neither a pool nor a garage, we subtract the probability of having either from 1. Therefore:

P( Neither ) = 1 - P( Pool or Garage )

= 1 - 0.91

= 0.09 or 9%

Answering Part c)

The probability of a home having a pool but no garage is calculated by taking the probability of having a pool and subtracting the probability of having both a pool and a garage:

P( Pool but no Garage ) = P( Pool ) - P( Pool and Garage )

= 0.29 - 0.13

= 0.16 or 16%

All but 1/13 of the students enrolled at a particular elementary school participated in "Family Fun Night" activities. If a total of 396 students were involved in the evening's activities, how many students attend the school?

Answers

Answer: 429

Step-by-step explanation:

If all but 1/13 of the students were involved in the evening's activities, that means that 12/13 of the students were involved.

To calculate the total number of students, first you do a simple rule of three to know the number of students that weren't involved in the activities:

If 12/13 of the students represent 396 students, then how many students are 1/13 of the students?

[tex]\frac{\frac{1}{13} *396}{\frac{12}{13} } =\frac{396}{12} =33[/tex]

Then, you add this number to the number of students that were involved in the activities:

[tex]396+33=429[/tex]

So, 429 is the number of students that attend the school.

You can also calculate it directly by doing the following rule of three:

If 12/13 of students represent 396 students, then how many students are 13/13 of the students?

[tex]\frac{\frac{13}{13} *396}{\frac{12}{13} } =\frac{396*13}{12} =\frac{5148}{12} =429[/tex]

Answer:429

Step-by-step explanation: let the total number of students in the elementary school be X

X - X/13 = 396

12X/13=396

X=396/*13/12

X=5148/12

X=429

The measurement of the height of 600 students of a college is normally distributed with a mean of
175 centimeters and a standard deviation of 5 centimeters.

What percent of students are between 180 centimeters and 185 centimeters in height?

12.5

13.5

34

68

Answers

Answer: Second Option

[tex]P(180<X <185)=13.5\%[/tex]

We know that the mean is:

[tex]\mu=175[/tex]

and the standard deviation is:

[tex]\sigma=5[/tex]

We are looking at the percentage of students between 180 centimeters and 185 centimeters in height.

This is:

[tex]P(180<X <185)[/tex]

We calculate the Z-score using the formula:

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

For [tex]X=180[/tex]

[tex]Z_{180}=\frac{180-175}{5}[/tex]

[tex]Z_{180}=1[/tex]

For [tex]X=185[/tex]

[tex]Z_{185}=\frac{185-175}{5}[/tex]

[tex]Z_{185}=2[/tex]

Then we look at the normal table

[tex]P(1<Z<2)[/tex]

[tex]P(1<Z<2)=P(Z<2)-P(Z<1)[/tex]

[tex]P(1<Z<2)=0.9772-0.8413[/tex]

[tex]P(1<Z<2)=0.135[/tex]

[tex]P(180<X <185)=13.5\%[/tex]

Note: You can get the same conclusion using the empirical rule

Look at the attached image for [tex]\mu+ 1\sigma <\mu <\mu + 2\sigma[/tex]

Students are collecting canned goods for a local food pantry. Last year 12 students were able to distribute 1000 flyers in nine hours. This year there are 15 students handing out the same number of flyers. How long should it take them?

Answers

Answer:

7.2 h

Step-by-step explanation:

The time required is inversely proportional to the number of students.

t = k/n or  

tn = k

Let t1  and n1 represent last year

and t2 and n2 represent this year. Then

t1n1 = t2n2

Data:

t1  = 9 h; n1 = 12 students

t2 = ?;    n2 = 15 students

Calculation:

9 × 12 = t2 × 15

   108 = 15t2

     t2 = 108/15 = 7.2 h

It will take the students 7.2 h to distribute the flyers.

Solve this differential Equation by using power series
y''-x^2y=o

Answers

We're looking for a solution

[tex]y=\displaystyle\sum_{n=0}^\infty a_nx^n[/tex]

which has second derivative

[tex]y''=\displaystyle\sum_{n=2}^\infty n(n-1)a_nx^{n-2}=\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n[/tex]

Substituting these into the ODE gives

[tex]\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^\infty a_nx^{n+2}=0[/tex]

[tex]\displaystyle\sum_{n=0}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0[/tex]

[tex]\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty(n+2)(n+1)a_{n+2}x^n-\sum_{n=2}^\infty a_{n-2}x^n=0[/tex]

[tex]\displaystyle2a_2+6a_3x+\sum_{n=2}^\infty\bigg((n+2)(n+1)a_{n+2}-a_{n-2}\bigg)x^n=0[/tex]

Right away we see [tex]a_2=a_3=0[/tex], and the coefficients are given according to the recurrence

[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_2=0\\a_3=0\\n(n-1)a_n=a_{n-4}&\text{for }n\ge4\end{cases}[/tex]

There's a dependency between terms in the sequence that are 4 indices apart, so we consider 4 different cases.

If [tex]n=4k[/tex], where [tex]k\ge0[/tex] is an integer, then

[tex]k=0\implies n=0\implies a_0=a_0[/tex]

[tex]k=1\implies n=4\implies a_4=\dfrac{a_0}{4\cdot3}=\dfrac2{4!}a_0[/tex]

[tex]k=2\implies n=8\implies a_8=\dfrac{a_4}{8\cdot7}=\dfrac{6\cdot5\cdot2}{8!}a_0[/tex]

[tex]k=3\implies n=12\implies a_{12}=\dfrac{a_8}{12\cdot11}=\dfrac{10\cdot9\cdot6\cdot5\cdot2}{12!}a_0[/tex]

and so on, with the general pattern

[tex]a_{4k}=\dfrac{a_0}{(4k)!}\displaystyle\prod_{i=1}^k(4i-2)(4i-3)[/tex]

If [tex]n=4k+1[/tex], then

[tex]k=0\implies n=1\implies a_1=a_1[/tex]

[tex]k=1\implies n=5\implies a_5=\dfrac{a_1}{5\cdot4}=\dfrac{3\cdot2}{5!}a_1[/tex]

[tex]k=2\implies n=9\implies a_9=\dfrac{a_5}{9\cdot8}=\dfrac{7\cdot6\cdot3\cdot2}{9!}a_1[/tex]

[tex]k=3\implies n=13\implies a_{13}=\dfrac{a_9}{13\cdot12}=\dfrac{11\cdot10\cdot7\cdot6\cdot3\cdot2}{13!}a_1[/tex]

and so on, with

[tex]a_{4k+1}=\dfrac{a_1}{(4k+1)!}\displaystyle\prod_{i=1}^k(4i-1)(4i-2)[/tex]

If [tex]n=4k+2[/tex] or [tex]n=4k+3[/tex], then

[tex]a_2=0\implies a_6=a_{10}=\cdots=a_{4k+2}=0[/tex]

[tex]a_3=0\implies a_7=a_{11}=\cdots=a_{4k+3}=0[/tex]

Then the solution to this ODE is

[tex]\boxed{y(x)=\displaystyle\sum_{k=0}^\infty a_{4k}x^{4k}+\sum_{k=0}^\infty a_{4k+1}x^{4k+1}}[/tex]

A company that makes thing-a-ma-bobs has a start up cost of $18263. It costs the company $1.14 to make each thing-a-ma-bob and the company charges $4.06 for each thing-a-ma-bob. Let x represent the number of thing-a-ma-bobs made. Write the cost function for this company. Write the revenue function for this company. Write the profit function for this company.

Answers

Answer:

c(x) = 18263 +1.14xr(x) = 4.06xp(x) = 2.92x -18263

Step-by-step explanation:

cost function

Cost for x tambs is the sum of start-up cost and per-unit cost multiplied by the number of units.

  c(x) = 18263 +1.14x

revenue function

Revenue from the sale of x tambs is the product of their price and the number sold.

  r(x) = 4.06x

profit function

Profit from the sale of x tambs is the difference between the revenue and cost:

  p(x) = r(x) -c(x)

  p(x) = 4.06x -(18263 +1.14x)

  p(x) = 2.92x -18263

The cost function, C(x), for the company is [tex]\( C(x) = 1.14x + 18263 \)[/tex] .The revenue function, R(x), for the company is [tex]\( R(x) = 4.06x[/tex] .The profit function, P(x), for the company is [tex]( P(x) = R(x) - C(x) = 4.06x - (1.14x + 18263) \)[/tex] .

To determine the profit function for the company, we need to calculate the total cost and total revenue for making and selling 'x' thing-a-ma-bobs, respectively, and then subtract the total cost from the total revenue.1. The cost function, C(x), is the sum of the variable cost (cost per unit times the number of units) and the fixed cost (start-up cost). Since it costs $1.14 to make each thing-a-ma-bob and the start-up cost is $18263, the cost function is:[tex]\[ C(x) = (\text{Cost per unit}) \times x + \text{Fixed cost} \]\[ C(x) = 1.14x + 182632[/tex]

. The revenue function, R(x), is the amount of money the company earns from selling 'x' thing-a-ma-bobs at $4.06 each. Therefore, the revenue function is[tex]:\[ R(x) = (\text{Selling price per unit}) \times x \][ R(x) = 4.06x3.[/tex]The profit function, P(x), is the revenue minus the cost. To find the profit for 'x' thing-a-ma-bobs, we subtract the cost function from the revenue function:[tex]\[ P(x) = R(x) - C(x) \]\[ P(x) = 4.06x - (1.14x + 18263) \]\[ P(x) = 4.06x - 1.14x - 18263 \]\[ P(x) = 2.92x - 18263 \]So, the profit function for the company is \( P(x) = 2.92x - 18263 \),[/tex]

where 'x' represents the number of thing-a-ma-bobs made and sold.

Find the unpaid balance on the debt. (Round your answer to the nearest cent.)

After 6 years of monthly payments on $190,000 at 3% for 25 years.

Answers

Answer:

unpaid balance is $156439.86

Step-by-step explanation:

present value= $190000

rate = 3% monthly = 3/12 = 0.25 % = 0.0025

time = 25 year  = 25 × 12 300

to find out

the unpaid balance

solution

first we calculate monthly payment

payment  = principal ( 1- (1+r)^(-t) /r )

and we put all value here

190000 = principal ( 1- (1+0.0025)^(-300) / 0.0025 )

principal = 190000 /  ( 1- (1+0.0025)^(-300) / 0.0025 )

principal = 901

so after 6 year unpaid is

present value  = principal ( 1- (1+r)^(-t) /r )

put here principal 901 and rate 0.0025 and time 300 - (year )

time = 300 - (6*12) = 300 - 96 = 228

payment   = 901 ( 1- (1+0.0025)^(-228) / 0.0025 )

payment  =  901 ( 713.63)

so unpaid balance is $156439.86

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