A curve passes through the point (0, 5) and has the property that the slope of the curve at every point P is twice the y-coordinate of P. What is the equation of the curve? (Use x as the independent variable.

Answer:

Proofs contained within the explanation.

Step-by-step explanation:

These induction proofs will consist of a base case, assumption of the equation holding for a certain unknown natural number, and then proving it is true for the next natural number.

a)

Proof

Base case:

We want to shown the given equation is true for n=1:

The first term on left is 2 so when n=1 the sum of the left is 2.

Now what do we get for the right when n=1:

[tex]\frac{1}{2}(1)(3(1)+1)[/tex]

[tex]\frac{1}{2}(3+1)[/tex]

[tex]\frac{1}{2}(4)[/tex]

[tex]2[/tex]

So the equation holds for n=1 since this leads to the true equation 2=2:

We are going to assume the following equation holds for some integer k greater than or equal to 1:

[tex]2+5+8+\cdots+(3k-1)=\frac{1}{2}k(3k+1)[/tex]

Given this assumption we want to show the following:

[tex]2+5+8+\cdots+(3(k+1)-1)=\frac{1}{2}(k+1)(3(k+1)+1)[/tex]

Let's start with the left hand side:

[tex]2+5+8+\cdots+(3(k+1)-1)[/tex]

[tex]2+5+8+\cdots+(3k-1)+(3(k+1)-1)[/tex]

The first k terms we know have a sum of .5k(3k+1) by our assumption.

[tex]\frac{1}{2}k(3k+1)+(3(k+1)-1)[/tex]

Distribute for the second term:

[tex]\frac{1}{2}k(3k+1)+(3k+3-1)[/tex]

Combine terms in second term:

[tex]\frac{1}{2}k(3k+1)+(3k+2)[/tex]

Factor out a half from both terms:

[tex]\frac{1}{2}[k(3k+1)+2(3k+2][/tex]

Distribute for both first and second term in the [ ].

[tex]\frac{1}{2}[3k^2+k+6k+4][/tex]

Combine like terms in the [ ].

[tex]\frac{1}{2}[3k^2+7k+4[/tex]

The thing inside the [ ] is called a quadratic expression.  It has a coefficient of 3 so we need to find two numbers that multiply to be ac (3*4) and add up to be b (7).

Those numbers would be 3 and 4 since

3(4)=12 and 3+4=7.

So we are going to factor by grouping now after substituting 7k for 3k+4k:

[tex]\frac{1}{2}[3k^2+3k+4k+4][/tex]

[tex]\frac{1}{2}[3k(k+1)+4(k+1)][/tex]

[tex]\frac{1}{2}[(k+1)(3k+4)][/tex]

[tex]\frac{1}{2}(k+1)(3k+4)[/tex]

[tex]\frac{1}{2}(k+1)(3(k+1)+1)[/tex].

Therefore for all integers n equal or greater than 1 the following equation holds:

[tex]2+5+8+\cdots+(3n-1)=\frac{1}{2}n(3n+1)[/tex]

//

b)

Proof:

Base case: When n=1, the left hand side is 1.

The right hand at n=1 gives us:

[tex]\frac{1}{4}(5^1-1)[/tex]

[tex]\frac{1}{4}(5-1)[/tex]

[tex]\frac{1}{4}(4)[/tex]

[tex]1[/tex]

So both sides are 1 for n=1, therefore the equation holds for the base case, n=1.

We want to assume the following equation holds for some natural k:

[tex]1+5+5^2+\cdots+5^{k-1}=\frac{1}{4}(5^k-1)[/tex].

We are going to use this assumption to show the following:

[tex]1+5+5^2+\cdots+5^{(k+1)-1}=\frac{1}{4}(5^{k+1}-1)[/tex]

Let's start with the left side:

[tex]1+5+5^2+\cdots+5^{(k+1)-1}[/tex]

[tex]1+5+5^2+\cdots+5^{k-1}+5^{(k+1)-1}[/tex]

We know the sum of the first k terms is 1/4(5^k-1) given by our assumption:

[tex]\frac{1}{4}(5^k-1)+5^{(k+1)-1}[/tex]

[tex]\frac{1}{4}(5^k-1)+5^k[/tex]

Factor out the 1/4 from both of the two terms:

[tex]\frac{1}{4}[(5^k-1)+4(5^k)][/tex]

[tex]\frac{1}{4}[5^k-1+4\cdot5^k][/tex]

Combine the like terms inside the [ ]:

[tex]\frac{1}{4}(5 \cdot 5^k-1)[/tex]

Apply law of exponents:

[tex]\frac{1}{4}(5^{k+1}-1)[/tex]

Therefore the following equation holds for all natural n:

[tex]1+5+5^2+\cdots+5^{n-1}=\frac{1}{4}(5^n-1)[/tex].

//

Answer:

Width 13, Length = 22

Step-by-step explanation:

Let L = length and W = width

Given, length is 9 more than width, ie.

L = W + 9  ----> eq 1

Also given that area = 286 square inches.

Area of rectangle = length x width , or

286 = L x W   -----> eq 2

2 equations, 2 unknowns => we can solve this system of equations by substitution.

Substitute eq 1 into eq 2,

286 = (W + 9) x W

286 = W² + 9W

W² + 9W - 286 = 0 (solve quadratic equation using your favorite method)

w = 13 or w = -22 (impossible since width cannot be negative)

hence W = 13 (substitute this back into equation 1)

L = 13 + 9 = 22

The width of the rectangle is 13 inches, and the length is 22 inches, obtained by solving the quadratic equation derived from the provided area and the relationship between the length and the width.

To find the length and width of a rectangle when given the area and the relationship between the length and the width, we can set up an algebraic equation. Let w represent the width and l represent the length of the rectangle. According to the problem, l = w + 9. The area of the rectangle, which is width times length, is given as 286 square inches, so our equation is w x (w + 9) = 286.

Solving this quadratic equation step by step:

Therefore, the width of the rectangle is 13 inches and the length is 22 inches.

Answer:

$0.60

Step-by-step explanation:

The costs are:

$0.15 + $0.25 = $0.40

He sells it for $1.25.

The value added is:

$1.25 - $0.40 = $0.85

The farmer adds a value of $0.60 to each pineapple by growing them and selling them to the convenience store.

The farmer obtains a revenue of $1.25 per pineapple, while the total production cost per pineapple is the sum of the seed, fertilizer, and forgone output costs, which amounts to $0.15 + $0.25 + $0.25 = $0.65. Therefore, the value added by the farmer to each pineapple is the revenue obtained from selling each pineapple minus the cost of producing each pineapple, which equals $1.25 - $0.65 = $0.60.

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Answer:

Step-by-step explanation:

When Riko left his house, Yuto was 5.25 miles along the path.

Average speed of Riko = 0.35 miles per minute

Average speed of Yuto = 0.25 miles per minute

First we will calculate the time in which Riko will catch Yuto on the track.

Relative velocity of Riko as compared to Yuto will be = velocity of Riko - velocity of Yuto

= 0.35 - 0.25

= 0.10 miles per minute

Now we this relative velocity tells that Riko is moving and Yuto is in static position.

By the formula,

Average speed = [tex]\frac{Distance}{Time}[/tex]

0.10 = [tex]\frac{5.25}{t}[/tex]

t = [tex]\frac{5.25}{0.1}[/tex]

t = 52.5 minutes

Now we know that Rico will catch Yuto in 52.5 minutes. Before this time he will be behind Yuto.

So duration of time in which Rico is behind Yuto will be 0 ≤ t ≤ 52.5

Here time can not be less than zero because time can not be with negative notation. It will always start from 0.

Answer:

Sample Response:

The solution means that Riko will be behind Yuto from the time she leaves the house, which corresponds to t = 0, until the time she catches up to Yuko after 52.5 minutes, which corresponds to t = 52.5. The reason that t cannot be less than zero is because it represents time, and time cannot be negative.

Answer:

5.25

Step-by-step explanation:

If CD is perpendicular to AB, a right angle is formed by the perpendicular lines at the base of the triangle. If we know that AB is 3 we can divided that by two to use half the triangle and create a right trianle with a base of 1.5 (half of 3) a height of rad 3 now all we have to do is use P-Thags to find AC which is the hypotenuse. After doing 1.5 squared + rad 3 squared = ac squared you will get the answer of AC = 5.25

Answer:

√(21) / 2

Step-by-step explanation:

The vertical line divides triangle ABC into two right angles which are congruent.

DB has length AB/2, which comes out to 3/2.  Regard DB as the base of one of the two right triangles.  If the height, CD, is √3, then by the Pythagorean Theorem |AC|² = (3/2)² + (√3)².

Thus, |AC| = √( (9/4) + 3 ), or  √( (9 + 12)/4 ), or √(21) / 2

[tex]\bf \textit{ellipse, horizontal major axis} \\\\ \cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2- b ^2} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]\bf 2x^2+8y^2=16\implies \cfrac{2x^2+8y^2}{16}=1\implies \cfrac{2x^2}{16}+\cfrac{8y^2}{16}=1\implies \cfrac{x^2}{8}+\cfrac{y^2}{2}=1 \\\\\\ \cfrac{(x-0)^2}{(\sqrt{8})^2}+\cfrac{(y-0)^2}{(\sqrt{2})^2}=1\qquad \begin{cases} a=\sqrt{8}\\ b=\sqrt{2}\\ c=\sqrt{a^2-b^2}\\ \qquad \sqrt{6} \end{cases}~\hfill \begin{array}{llll} \stackrel{\textit{vertices}}{(\pm \sqrt{8},0)}\qquad \stackrel{\textit{foci}}{(\pm \sqrt{6},0)}\\\\ ~\hfill \stackrel{\textit{center}}{(0,0)}~\hfill \end{array}[/tex]

Answer:

Step-by-step explanation:

2x² + 8y² = 16

divide both sides of equation by the constant

2x²/16 + 8y²/16 = 16/16

x²/8 + y²/2 = 1

x² has a larger denominator than y², so the ellipse is horizontal.

General equation for a horizontal ellipse:

  (x-h)²/a² + (y-k)²/b² = 1

with

  a² ≥ b²

  center (h,k)

  vertices (h±a, k)

  co-vertices (h, k±b)

  foci (h±c, k), c² = a²-b²

Plug in your equation, x²/8 + y²/2 = 1.

(x-0)²/(2√2)² + (y-0)²/(√2)² = 1

h = k = 0

a = 2√2

b = √2

c² = a²-b² = 6

c = √6

center (0,0)

vertices (0±2√2,0) = (-2√2, 0) and (2√2, 0)

co-vertices (0, 0±√2) = (0, -√2) and (0, √2)

foci (0±√6, 0) = (-√6, 0) and (√6, 0)

Answer:

  TRUE

Step-by-step explanation:

They are called independent and consistent.

_____

If they describe the same line, they are dependent. If they describe parallel lines, they are inconsistent.

Answer:

First problem: a (2,0)

Second problem: b. none of these; the answer is (4, 5/3) which is not listed.

Third problem: b. none of the above; there are no holes period.

Step-by-step explanation:

First problem:  The hole is going to make both the bottom and the top zero.

So I start at the bottom first.

[tex]x^2-3x+2=0[/tex]

The left hand expression is factorable.

Since the coefficient of [tex]x^2[/tex] is 1, you are looking for two numbers that multiply to be 2 and add to be -3.

Those numbers are -2 and -1 since (-2)(-1)=2 and -2+(-1)=-3.

The factored form of the equation is:

[tex](x-2)(x-1)=0[/tex].

This means x-2=0 or x-1=0.

We have to solve both equations here.

x-2=0

Add 2 on both sides:

x=2

x-1=0

Add 1 on both sides:

x=1

Now to determine if x=2 or x=1 is a hole, we have to see if it makes the top 0.

If the top is zero when you replace in 2 for x, then x=2 is a hole.

If the top is zero when you replace in 1 for x, then x=1 is a hole.

Let's do that.

[tex]x^2-4x+4[/tex]

x=2

[tex]2^2-4(2)+4[/tex]

[tex]4-8+4[/tex]

[tex]-4+4[/tex]

[tex]0[/tex]

So we have a hole at x=2.

[tex]x^2-4x+4[/tex]

x=1

[tex]1^2-4(1)+4[/tex]

[tex]1-4+4[/tex]

[tex]-3+4[/tex]

[tex]1[/tex]

So x=1 is not a hole, it is a vertical asymptote.  We know it is a vertical asymptote instead of a hole because the numerator wasn't 0 when we plugged in the x=1.

So anyways to find the point for which we have the hole, we will cancel out the factor that makes us have 0/0.

So let's factor the denominator now.

Since the coefficient of [tex]x^2[/tex] is 1, all we have to do is find two numbers that multiply to be 4 and add up to be -4.

Those numbers are -2 and -2 because -2(-2)=4 and -2+(-2)=-4.

[tex]f(x)=\frac{(x-2)(x-2)}{(x-2)(x-1)}=\frac{x-2}{x-1}[/tex]

So now let's plug in 2 into the simplified version:

[tex]f(2)=\frac{2-2}{2-1}=\frac{0}{1}=0[/tex].

So the hole is at x=2 and the point for which the hole is at is (2,0).

a. (2,0)

Problem 2:

So these quadratics are the same kind of the ones before. They all have coefficient of [tex]x^2[/tex] being 1.

I'm going to start with the factored forms this time:

The factored form of [tex]x^2-3x-4[/tex] is [tex](x-4)(x+1)[/tex] because -4(1)=-3 and -4+1=-3.

The factored form of [tex]x^2-5x+4[/tex] is [tex](x-4)(x-1)[/tex] because -4(-1)=4 and -4+(-1)=-5.

Look at [tex]\frac{(x-4)(x+1)}{(x-4)(x-1)}[/tex].

The hole is going to be when you have 0/0.

This happens at x=4 because x-4 is 0 when x=4.

The hole is at x=4.

Let's find the point now. It is (4,something).

So let's cancel out the (x-4)'s now.

[tex]\frac{x+1}{x-1}[/tex]

Plug in x=4 to find the corresponding y:

[tex]\frac{4+1}{4-1}{/tex]

[tex]\frac{5}{3}[/tex]

The hole is at (4, 5/3).

Third problem:

[tex]x^2-4x+4[/tex] has factored form [tex](x-2)(x-2)[/tex] because (-2)(-2)=4 and -2+(-2)=-4.

[tex]x^2-5x+4[/tex] has factored form [tex](x-4)(x-1)[/tex] because (-4)(-1)=4 and -4+(-1)=-5.

There are no common factors on top and bottom.  You aren't going to have a hole.  There is no value of x that gives you 0/0.

Answer:

  D.)  (14, 0); the time it takes for the bird to reach the ground

Step-by-step explanation:

The attached graph shows a plot of the table values and the two offered solution options.

The dependent variable in this scenario is the bird's height above the ground. When that is zero, the bird is on the ground. This fact makes choices B and C seem ridiculous.

We note from the table and graph that the bird is on a path that decreases in height by 3 feet every 2 seconds. If the bird continues that rate of descent, it will reach the ground after 14 seconds.

That is, its (time, height) pair will be (14, 0), matching choice D.

_____

Choosing any answer to this question requires making assumptions that are inconsistent with real-world bird behavior. At least, the problem statement should say what assumptions are applicable.

Answer: D.) (14, 0); the time it takes for the bird to reach the ground

Step-by-step explanation:

The total number of phone numbers that can be made is 1,200,000.

To find the number of possible phone numbers given the conditions:

We can calculate as follows:

Therefore, the total number of phone numbers that can be made is [tex]1 x 3 x 4 x 10^7[/tex] = 1,200,000.

Answer: 210

Step-by-step explanation:

55×12=660-500=160+50(down payment)=210

The total amount(interset+down payment) is $210 if the Rahm used a payment plan to purchase wood for a home project. The wood he bought cost $500.

Paying down any outstanding debt, or occasionally more than one obligation, by consolidation into a structured payment schedule is referred to as a payment plan.

We have the wood he bought cost $500.

Here some data are missing, so we are assuming the monthly payment is $55 and duration is 1 year

= 55×12   (1 year = 12 months)

= $660

= 600 – 500

= $160

Down payment = $50

Total amount = 160+50 = $210

Thus, the total amount(interset+down payment) is $210 if the Rahm used a payment plan to purchase wood for a home project. The wood he bought cost $500.

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Answer:

Step-by-step explanation:

a) 189 km/h = (189 km/h)(1 h/(3600 s))(1000 m/km) = 189/3.6 m/s = 52.5 m/s

__

b) In 5 seconds, the parachutist will fall ...

  (5 s)(52.5 m/s) = 262.5 m

Answer:

If you meant [tex]9^x+4=11[/tex], then the answer is approximately 0.866.

If you meant [tex]9^{x+4}=11[/tex], then the answer is approximately -2.909 which looks like what you meant based on the choices.

Step-by-step explanation:

[tex]9^x+4=11[/tex]

First step is to get the exponential part by itself. The part that has the variable exponent which is the [tex]9^x[/tex] term.

To do this we need to subtract 4 on both sides:

[tex]9^x=11-4[/tex]

Simplify:

[tex]9^x=7[/tex]

The equivalent logarithmic form is:

[tex]\log_9(7)=x[/tex]

I always say to myself the logarithm is the exponent that is how I know what to put opposite the side containing the log.

Anyways if you don't have options for computing [tex]\log_b(a)[/tex] in your calculator you need to use the change of base formula.

[tex]\frac{\log(7)}{\log(9)}=x[/tex]

So [tex]x \approx 0.8856[/tex]

I don't see this as a choice so maybe you actually meant the following equation:

[tex]9^{x+4}=11[/tex]

Let's see if this is the correct interpretation.

So the exponential part is already isolated.

So we just need to put in the equivalent logarithmic form:

[tex]\log_9(11)=x+4[/tex]

Now we subtract 4 on both sides:

[tex]\log_9(11)-4=x[/tex]

Again if you don't have the option for computing [tex]\log_b(a)[/tex] in your calculator, you will have to use the change of base formula:

[tex]\frac{\log(11)}{\log(9)}-4=x[/tex]

[tex]x \approx -2.909[/tex]

i know im late, but the answer is b. -2.909 :)

Yes, y is a function of x in the given equation x = y³ - 10.

In the equation x = y³ - 10, y is indeed a function of x. To determine whether y is a function of x, we need to check if each x-value corresponds to a unique y-value. In this case, for every x, there exists only one corresponding y. Let's consider the equation step by step.

The given equation x = y³ - 10 implies that y³ = x + 10. To solve for y, we take the cube root of both sides, yielding y = [tex](x + 10)^(^1^/^3^)[/tex]. This expression defines y explicitly in terms of x, confirming that for every x, there is a unique y. Thus, the equation satisfies the criteria for a function.

Examining the nature of the equation further, we observe that the term[tex](x + 10)^(^1^/^3^)[/tex] represents a real-valued function. The cube root of any real number is a single-valued function, ensuring that y is indeed uniquely determined by x. Therefore, we can confidently conclude that y is a function of x in the given equation x = y³ - 10.

Answer:

2.5 mg

Step-by-step explanation:

Substitute the givens into the equation. A is the initial amount, 16 mg. H is the halflife, 8 days. T is the time in days that has passed, 16 days. So we get P(t)= 10(1/2)^(16/2). This ends up being 10(1/2)^2. 1/2^2 is 1/4. 10(1/4)=2.5 mg

To understand how to use the equation P(t) = A(1/2)^(t/h), let's break down each part of the formula and see how it applies to a real-world situation.
P(t): This represents the remaining amount of the substance at time t, where t is measured in days.
A: This is the initial amount of the substance before any decay has started.
(1/2): This factor represents the principle of half-life, which, in this context, means that the substance is reduced to half its previous amount after each half-life period passes.
t: This is the time that has passed, measured in days.
h: This is the half-life of the substance, which is the amount of time it takes for half of the substance to decay.
The half-life formula can be used to calculate the amount of substance that will remain after a certain amount of time has passed. Here is how you use it:
1. Start by determining the initial amount A of the substance. This is how much of the substance you begin with.
2. Determine the half-life h of the substance, which is usually provided by scientific data or an experiment.
3. Choose the time period t that you are interested in. This is how many days from the start time you want to know the remaining amount of the substance for.
4. Plug the values of A, h, and t into the formula P(t) = A(1/2)^(t/h).
5. Calculate (1/2)^(t/h). This requires you to raise (1/2) to the power of the fraction t/h. This fraction is the number of half-lives that have passed in the time period t.
6. Multiply the initial amount A by the result from step 5 to get P(t), the amount of the substance that remains after t days.
Let's go through an example to make it clear:
Example:
If the initial amount A is 100 grams and the half-life h is 10 days, how much of the substance will remain after 20 days?
Using the formula:
1. A = 100 grams (initial amount)
2. h = 10 days (half-life)
3. t = 20 days (time passed)
Plug the values into the formula:
P(t) = A(1/2)^(t/h)
P(20) = 100(1/2)^(20/10)
Calculate the exponent:
(1/2)^(20/10) = (1/2)^2 = 1/4
Multiply the initial amount by the result of the exponent:
P(20) = 100 * 1/4
P(20) = 25 grams
So after 20 days, 25 grams of the substance would remain.

The 3 known points of the parabola are the origin (0,0) and the bottom left and right points ( -5,-50) (5,-50)

The general form of a parabola that goes through the origin is:  

y = ax^2  

We need to solve for a:

-50 = a(5)^2  

-50 = 25a  

a = -50/25

a = -2

Now the equation of the parabola is y = (-2)x^2.

To solve for the vertical clearance (y) 2 meters from the edge

Subtract the distance from the edge 2 meters from the original edge which is 5 meters, 5-2 = 3.

Now replace x with 3 and solve for y:

y = (-2)3^2

y = -2(9)

y = -18

The clearance would then 50 - 18 = 32 meters.

Answer:

D

Step-by-step explanation:

Using the product to sum formula

• 2cosAcosB = cos(A+B) + cos(A - B)

note that cos(- 3m) = cos 3m, hence

cos5m cos(- 3m)

= cos 5m cos3m ← A = 5m and B = 3m

= [tex]\frac{1}{2}[/tex] ( 5m + 3m) + cos(5m - 3m) ]

= [tex]\frac{1}{2}[/tex] cos 8m + [tex]\frac{1}{2}[/tex] cos 2m

The expression of cos5mcos(-3m) as a sum or difference results in  1/2cos2m + 1/2cos8m.

To express cos5mcos(-3m) as a sum or difference, we can use the trigonometric identity cos(A)cos(B) = 1/2[cos(A+B) + cos(A-B)].

Applying this identity to the given expression, we get 1/2[cos(5m - 3m) + cos(5m + 3m)] = 1/2[cos(2m) + cos(8m)].

Therefore, the given expression cos5mcos(-3m) can be simplified to 1/2cos2m + 1/2cos8m.

Using the principle of odds representation in probability, the odds that a card chosen is not spade ls 3 : 1

Not spade = (total number of cards - number of spades)

Not spade = (52 - 13) = 39

The odds of not picking a spade can be represented thus :

Therefore, the odds of not choosing a spade is 3 : 1

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In a deck of 52 cards, the odds of not drawing a spade are 3:1, meaning that for every three times one pulls a non-spade card, one time will yield a spade.

In a standard deck of 52 playing cards, there are 13 spades. Hence, the number of successful outcomes for picking a non-spade card is 52 - 13 = 39. The number of unsuccessful outcomes for not picking a spade is 13. So, the odds of not picking a spade (successful outcome) to picking a spade (unsuccessful outcome) are 39:13, which can be reduced to a simpler ratio of 3:1. Therefore, for every three times one pulls a non-spade card, one time will yield a spade.

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Surveying people coming out of a concert because it will be all people who like the concert they were just attending.

Answer: leaving a concert

Step-by-step explanation:

most of the people leaving the concert probably really like that kind of music so it would be biased to survey them

Answer:

A

Step-by-step explanation:

To find the coordinates of the vertices , treat the inequalities as equations and solve them in pairs, simultaneously, that is

x + y = 9 → (1)

y - 2x = - 21 → (2) ← add 2x to both sides

y = - 21 + 2x → (3)

Substitute y = - 21 + 2x into (1)

x - 21 + 2x = 9

3x - 21 = 9 ( add 21 to both sides )

3x = 30 ( divide both sides by 3 )

x = 10

Substitute x = 10 into (3)

y = - 21 + 20 = - 1

Coordinates of vertex = (10, - 1 )

---------------------------------------------------------------------

y - 2x = - 21 → (1)

y = - 4x + 15 → (2)

Substitute y = - 4x + 15 into (1)

- 4x + 15 - 2x = - 21

- 6x + 15 = - 21 ( subtract 15 from both sides )

- 6x = - 36 ( divide both sides by - 6 )

x = 6

Substitute x = 6 into (2)

y = - 24 + 15 = - 9

Coordinates of vertex = (6, - 9 )

------------------------------------------------------------------------

x + y = 9 → (1)

y = - 4x + 15 → (2)

Substitute y = - 4x + 15 into (1)

x - 4x + 15 = 9

- 3x + 15 = 9 ( subtract 15 from both sides )

- 3x = - 6 ( divide both sides by - 3 )

x = 2

Substitute x = 2 into (2)

y = - 8 + 15 = 7

Coordinates of vertex = (2, 7 )

Answer:

Choice d.

Step-by-step explanation:

Consistent means the linear functions will have at least one solution.

Independent means the it will just that one solution.

Dependent means the system will have infinitely many solutions.

So we are looking for a pair of equations that consist of the same line.

y=mx+b is slope-intercept form where m is the slope and b is the y-intercept.

If m and b are the same amongst the pair, then that pair is the same line and the system is consistent and dependent.

If m is the same and b is different amongst the pair, then they are parallel and the system will be inconsistent (have no solution).

If m is different, then the pair will have one solution and will be consistent and independent.

---------------------------

Choice A:

First line has m=1 and b=4.

Second line has m=1 and b=-4.

These lines are parallel.

This system is inconsistent.

Choice B:

First line is x+y=4 and it isn't in y=mx+b form.

Second line is 2x+2y=6 and it isn't in y=mx+b form.

These lines do have the same form though. This is actually standard form.

If you divide second equation by 2 on both sides you get:  x+y=3

The lines are not the same.

We are looking for the same line.

Now we could go ahead and determine to call this system.

If x+y has value 4 then how could x+y have value 3.  It is not possible. There is no solution. These lines are parallel.

Need more convincing. Let's put them into slope-intercept form.

x+y=4

Subtract x on both sides:

y=-x+4

m=-1 and b=4

x+y=3

Subtract x on both sides:

y=-x+3

m=-1 and b=3.

The system is inconsistent because they are parallel.

Choice c:

I'm going to go ahead in put them both in slope-intercept form:

3x+y=3

Subtract 3x on both sides:

y=-3x+3

So m=-3 and b=3

2y=6x+6

Divide both sides by 2:

y=3x+3

So m=3 and b=3

The m's are different so this system will by consistent and independent.

Choice d:

The goal is the same. Put them in slope-intercept form.

4x-2y=6

Divide both sides by -2:

-2x+y=-3

Add 2x on both sides:

y=2x-3

m=2 and b=-3

6x-3y=9

Divide both sides by -3:

-2x+y=-3

Add 2x on btoh sides:

y=2x-3

m=2 and b=-3

These are the same line because they have the same m and the same b.

This system is consistent and dependent.

Answer:

  26 minutes on the bicycle

Step-by-step explanation:

Let t and b represent the number of minutes on the treadmill and bicycle, respectively. The problem statement tells us two ways to relate these values.

  t + b = 60 . . . . . Amy exercises a total of 60 minutes (1 hour)

  9t +6.5b = 475 . . . . . she burns a total of 475 calories

Since we want to find the value of "b", we can use the first equation to write an expression that will substitute for t.

  t = 60 -b

  9(60 -b) +6.5b = 475

  540 -2.5b = 475 . . . . . . simpify

  -2.5b = -65 . . . . . . . . . . subtract 540

  b = 26 . . . . . . . . . . . . . . . divide by -2.5 (equivalently, multiply by -2/5)

Amy spends 26 minutes exercising on the bicycle.

Setting up a system of linear equations can help us find the time spent on each exercise. Form two equations based on the given information: total time spent exercising and total calories burned. Solve these equations to find the time spent on each activity.

To solve these kinds of problems, it is helpful to use a system of linear equations. We need to find two expressions for the total calories burned, based on the different exercises Amy does, and then solve for the time she spends on each exercise.

Step 1: Identify the variables. Let's denote the time Amy spends jogging as J (in minutes) and the time she spends bicycling as B (also in minutes).

Step 2: Write two equations based on the information given. We know:

Step 3: Solve for B in the first equation (B = 60 - J) and substitute this into the second equation. Simplifying will give J, the number of minutes Amy spends jogging. Substituting J back into the first equation will give B, the number of minutes spent bicycling.

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Answer:

B 1 1/7

Step-by-step explanation:

- 9 2/7 - ( -10 3/7)

We know subtracting a negative is like adding

- 9 2/7 +10 3/7

10 3/7 - 9 2/7

Putting this vertical

10  3/7

-9  2/7

-----------------

1   1/7

Answer:

1/17

Step-by-step explanation:

Answer:

64 oz

Step-by-step explanation:

Check the picture below.

notice, all three sides in the triangle are of different lengths, thus is an scalene triangle.

The correct answer is C. on edg

Answer:

The correct answer is C. on edg

Step-bexplanation:

Answer with explanation:

Total number of Children = x Children

Total number of chocolates distributed = y Chocolates

If there are x children, then total number of chocolate got by children= 5 x Chocolate

Number of chocolate got by Adults= 20

Let total number of adults be z.

Expressing the statement in terms of Equation

 y = 5 x + 20 z

[tex]z=\frac{y-5 x}{20}[/tex]

⇒Number of Adults =z

       [tex]=\frac{y-5 x}{20}[/tex]

715 calculators will be defective, from a sample of 13200.

Calculation using the unitary method:-

In a sample of 240 = 13 calculators are found defective.

so, in a sample of 1 =13/240 calculator is defective.

 ∴in a sample of 13200 = 13/240*13200 calculators are defective.

                                        715 calculators are defective (answer)

Problem-solving is the act of defining a problem; figuring out the purpose of the trouble; identifying, prioritizing, and selecting alternatives for an answer; and imposing an answer.

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