What is the value of cosA ?

Answer: The value of f is 1.

Step-by-step explanation:

Since we have given that

[tex]-f+2+4f=8-3f[/tex]

We need to find the value of f :

1) First we gather the like terms together:

2) Solving the like terms

3)find the value of f.

[tex]-f+2+4f=8-3f\\\\3f+2=8-3f\\\\3f+3f=8-2\\\\6f=6\\\\f=\dfrac{6}{6}\\\\f=1[/tex]

Hence, the value of f is 1.

The solution to the equation -f + 2 + 4f = 8 - 3f is f = 1.

To solve the equation -f + 2 + 4f = 8 - 3f, we can simplify and solve for f:

Combine like terms on the left side:

-f + 4f + 2 = 8 - 3f

Simplify: 3f + 2 = 8 - 3f

Add 3f to both sides:

3f + 3f + 2 = 8 - 3f + 3f

Simplify: 6f + 2 = 8

Subtract 2 from both sides:

6f + 2 - 2 = 8 - 2

Simplify: 6f = 6

Divide both sides by 6:

6f/6 = 6/6

Simplify: f = 1

Therefore, the solution to the equation -f + 2 + 4f = 8 - 3f is f = 1.

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$2112 you just have to do 528 x 4 = 2112

The company's total profit over four months, if sustaining a consistent loss of $528 per month, would be -$2,112.

If a small company had a profit of -$528 in January, and it continues to make the same profit (which is really a loss) each month for four months, we can calculate the total profit (total loss in this case) for those four months by multiplying the monthly profit by four.

So, the calculation would be:

Monthly Profit x Number of Months = Total Profit

(-$528) x 4 = -$2,112

Therefore, the company's total profit after four months would be -$2,112. This means the company would have a loss of $2,112 total over those four months.

Answer:   (i)  The unit rate at which online jewelry sales have been increasing is 1.33 billion dollars per year.

(ii)  The online jewelry sales in 2019 will be 191.7 billion dollars.

Step-by-step explanation:

In 2003, sales were approximately 2 billion dollars and in 2010, they were approximately 14.8 billion dollars.

(i)  If [tex]x[/tex] is the number of years after 2003 and [tex]y[/tex] is the amount of sales....

then the equation will be:   [tex]y= ab^x[/tex] , where [tex]a[/tex] is the initial amount and [tex]b[/tex] is the growth rate.

for 2003,  [tex]x=0[/tex] and for 2010,  [tex]x=7[/tex]

So, the two points in form of (x, y) will be:   [tex](0,2)[/tex] and [tex](7,14.8)[/tex]

Now plugging these two points int the above equation....

[tex]2= ab^0\\ \\ a= 2\\ \\ and\\ \\ 14.8=ab^7\\ \\ 14.8=2*b^7\\ \\ b^7=7.4\\ \\b= \sqrt[7]{7.4}=1.3309.... \approx 1.33[/tex]

Thus, the online jewelry sales have been increasing at a rate of 1.33 billion dollars per year.

(ii)   As we got [tex]a=2[/tex] and [tex]b=1.33[/tex], so the equation will be now:  [tex]y= 2(1.33)^x[/tex]

For the year 2019, the value of  [tex]x[/tex] will be: (2019-2003) = 16

So plugging [tex]x=16[/tex] into the above equation, we will get.....

[tex]y=2(1.33)^16\\ \\ y=191.7150... \approx 191.7[/tex]

(Rounded to the nearest tenth)

Thus, the online jewelry sales in 2019 will be 191.7 billion dollars.

y = - 2x + 12

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

y = - 2x + 7 is in this form with slope m = - 2

Parallel lines have equal slopes, thus

y = - 2x + c is the partial equation.

To find c, substitute (3, 6 ) into the partial equation

6 = - 6 + c ⇒ c = 6 + 6 = 12

y = - 2x + 12 ← equation of parallel line

(x + 2 )² + (y - 4)² = 36 ← is the equation of the circle

the equation of a circle in standard form is (x - a)² + (y - b)² = r²

where (a, b) are the coordinates of the centre and r is the radius

here (a, b ) =(- 2, 4) and r = 6

(x + 2)² + ( y - 4 )² = 36

We are required to find the least common denominator for the fractions 1/2 and 2/3

The least common denominator for the fractions 1/2 and 2/3 3/6 and 4/6

Given:

1/2 and 2/3

Find the lowest common multiples of the denominators 2 and 3

2 = 2, 4, 6, 8, 10, 12, 14, 16, 18, 30

3 = 3, 6, 9, 12, 15, 18, 21

The lowest common multiples of the denominators 2 and 3 is 6

Check:

3/6 and 4/6

= 1/2 and 2/3

Therefore, the least common denominator for the fractions 1/2 and 2/3 is 3/6 and 4/6

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The least common denominator for the fractions 1/2 and 2/3 is 6, which is the least common multiple of their denominators. Fractions are converted to have this common denominator before performing operations such as addition or comparison.

The least common denominator (LCD) for the fractions 1/2 and 2/3 is the smallest number that both denominators can divide into without leaving a remainder. To find the LCD, we look for the least common multiple (LCM) of the two denominators. In this case, the denominators are 2 and 3. The LCM of 2 and 3 is 6, because 6 is the smallest number that both 2 and 3 can divide into evenly. Therefore, the least common denominator for 1/2 and 2/3 is 6.

To express the fractions with the common denominator, you would convert 1/2 to 3/6 and 2/3 to 4/6. This allows you to add or compare the fractions directly.

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Rectangles are similar figures, thus if scaled copies of each other then the ratios of corresponding sides must be equal

compare ratios of lengths and widths

rectangles A and B

k = [tex]\frac{12}{15}[/tex] = [tex]\frac{4}{5}[/tex] ← ratio of lengths

k = [tex]\frac{8}{10}[/tex] = [tex]\frac{4}{5}[/tex] ← ratio of widths

scale factors are equivalent, hence rectangle A is a scaled copy of B

rectangles C and B

k = [tex]\frac{15}{30}[/tex] = [tex]\frac{1}{2}[/tex] ← ratio of lengths

k = [tex]\frac{10}{15}[/tex] = [tex]\frac{2}{3}[/tex] ← ratio of width

scale factors (k ) are not equal, hence C is not a scaled copy of B

rectangles A and C

k = [tex]\frac{30}{12}[/tex] = [tex]\frac{5}{2}[/tex] ← ratio of lengths

k = [tex]\frac{15}{8}[/tex] ← ratio of widths

the scale factors are not equal hence A is not a scaled copy of C

For two rectangles, one of length L and width W, and other of length L' and width W', the second is a rescale of the first one only if exists a real number k such that:

L' = k*L

W' = k*W

Here we know:

Let's see if rectangle A is a scaled copy of rectangle B.

To see this, we just must see if the quotients between the lengths and between the widths are equal:

15/12 = 1.25

10/8 =  1.25

Then yes, rectangle A is a rescaled copy of rectangle B, and the scale factor is k = 1.25

Is rectangle B a rescaled copy of rectangle A?

Obviously yes. The scale factor will be the inverse of the previous one, we will get:

k = 1/1.25 = 0.8

How we do know that rectangle C is not a scaled copy of rectangle B?

Because the length of C is twice the length of B, but the width of C is not twice the width of B.

Is rectangle A a scaled copy of rectangle C?

No, as we already see that rectangle C is not a rescaled copy of rectangle B, and we know that rectangle A is a rescaled copy of rectangle B.

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16. Vertical angles are the ones bounded by the same lines and have the same vertex, but that have no sides in common. Pairs 1 and 3 or 2 and 4 are vertical angles.

17. The diagram shows the sum of the three angles makes a right angle (90°). Write that as an equation:

... x° + 2x° + 15° = 90°

Solve the equation in the usual manner: collect terms, add the opposite of the unwanted constant on the left, divide by the coefficient of x.

... 3x° +15° = 90°

... 3x° = 75°

... x = 25

18. You may notice that this problem follows the same pattern as the one of 17. We add the constituent angles to make the whole right angle. Here, you have some follow-on effort to find ∠BDC after you find x.

... (-3x+20)° + (-2x+55)° = 90°

... -5x +75 = 90 . . . . . . . collect terms, divide by °

... -5x = 15 . . . . . . . . . . . subtract 75

... x = -3 . . . . . . . . . . . . . divide by the coefficient of x

Now we can find ∠BDC.

... ∠BDC = (-3x+20)° = (-3(-3)+20)°

... ∠BDC = 29°

1 [tex]\frac{3}{4}[/tex] gallons per gour

divide number of gallons used by running time

4 [tex]\frac{3}{8}[/tex]÷ 2 [tex]\frac{1}{2}[/tex]

change mixed numbers to improper fractions

[tex]\frac{35}{8}[/tex] ÷ [tex]\frac{5}{2}[/tex]

leave the first fraction, change division to multiplication, turn the second fraction upside down

= [tex]\frac{35}{8}[/tex] × [tex]\frac{2}{5}[/tex]

= ( 35 × 2 ) / (8 × 5 ) = [tex]\frac{7}{4}[/tex] = 1 [tex]\frac{3}{4}[/tex]

Answer:

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

Step-by-step explanation:

Solving the equation mean finding the value of x

Equation given is:

[tex]-3x+1+10x=x+4[/tex]

Now what we need to do is take the values with x in it to the left side of the equation and the other numbers to the right side of the equation.

[tex]-3x+10x-x=4-1[/tex]

Now simplify values with x and the numbers.

[tex]6x=3[/tex]

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

Therefore [tex]x=[tex]\frac{1}{2}[/tex][/tex]

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

simplify the left side by collecting like terms

7x + 1 = x + 4 ( subtract x from both sides )

6x + 1 = 4 ( subtract 1 from both sides )

6x = 3 ( divide both sides by 6 )

x = [tex]\frac{3}{6}[/tex] = [tex]\frac{1}{2}[/tex]

Answer: Procedure are given below :

Step-by-step explanation:

Since we have given that

Pay rate = $21.00

if he marked up by 78.7% , then

[tex]78.7\%\text{ of }21\\\\=\frac{78.7}{100}\times 21\\\\=16.527[/tex]

So, our pay rate becomes

[tex]\$21+\$16.527=\$37.527[/tex]

Now, he made a 10% profit,

[tex]10\%\text{ of }37.527\\\\=\frac{10}{100}\times 37.527\\\\=\$3.7527[/tex]

So, pay rate becomes

[tex]\$37.527+\$3.7527\\\\=\$41.2797[/tex]

which approximately $41.28

The distance formula is:  Distance = Rate x Time.

We know the distance: 2000 miles.

We know the rates: 560 and 500 mph.

We need to solve for time:

2000 = (560 + 500) * T

2000 = 1060 *T

T = 2000 / 1060

T = 1.89 hours ( 1 hour 53 minutes)

They left at noon:

12:00 pm + 1 hour and 53 minutes = 1:53 pm.

The two planes are [tex]2000[/tex] miles apart at [tex]1:53[/tex] p.m.

Distance [tex]= 2000[/tex] miles

When both are traveling in opposite directions, speeds are added.

So, net speed [tex]= (560+500) = 1060[/tex] mph.

[tex]Speed = \frac{Distance}{Time}[/tex]

[tex]Time = \frac{Distance}{Speed}[/tex]

[tex]Time = \frac{2000}{1060}[/tex]

[tex]Time = 1.89[/tex] hour  

or  

Time [tex]= 1[/tex] hour [tex]53[/tex] minutes.

They leave at noon:

[tex]12:00[/tex] p.m [tex]+ 1[/tex] hour [tex]53[/tex] minutes [tex]= 1:53[/tex] p.m

So, the two planes are [tex]2000[/tex] miles apart at [tex]1:53[/tex] p.m.

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x^3−4x^2−3x+18=0

(x - 3)^2 (x + 2) = 0

Choice B

A

since x = 3 is a zero then (x - 3) is a factor

dividing x³ - 4x² - 3x + 18 by (x - 3) gives

x³ - 4x² - 3x + 18 = (x - 3)(x² - x - 6 ) = 0,

(x - 3)(x - 3 )(x + 2 ) = 0

(x + 2 )(x - 3)² = 0 → A

Any arithmetic operation on polynomials except division* will result in a polynomial. Appropriate choices are ...

A. (x^2+2) (x-1)

C. (x+1) plus (X^2-3X -2)

D. (X +5)-(3X +6)

E. (4/8)x . . . . . but not if you mean 4/(8x)

_____

* For division, the result may be a polynomial. In the specific example given here for B, it is not.

The term closure in mathematics refers to the property of a set, in this case, polynomials, where doing certain operations (addition, subtraction, multiplication) with any numbers from the set always yields a number within the set. The options demonstrating closure in a polynomial are A, C, D, E, all but option B, which implies division.

The term 'closure' in mathematics refers to the property of a set under an operation where the operation performed on any numbers in the set always produces a number that is also in that set. In the case of polynomials, the operations could be addition, subtraction, or multiplication.

The expressions that demonstrate closure in a polynomial for addition, subtraction, and multiplication are all but option B. Option B, '(x^2+2) over (X -1)', isn't representative of closure in a polynomial as it implies a division operation. Polynomial closure doesn't include division because it can produce numbers outside the original set.

So, the correct answers are A.(x^2+2) (x-1), C.(x+1) plus (X^2-3X -2), D.(X +5)-(3X +6) and E.4/8x as they illustrate closure by either multiplication of two polynomials, addition of two polynomials, or subtraction of two polynomials.

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The exponents differ by 5, meaning the numbers differ by a factor of 10⁵.

9.125×10² = 100,000×9.125×10⁻³

The latter is 100,000 times the former

In 4 minutes, 1/15 hour, the robbers have a (96 mi/h)·(1/15 h) = 6.4 mile head start.

The cops must travel 65 miles in the time the robbers travel 48.6 miles. Hence their speed must be 65/48.6 times that of the robbers, or about

... 65/48.6×96 mi/h ≈ 128.4 mi/h

_____

The chase will be over after (55 mi)/(96 mi/h) = 34 minutes 22.5 seconds from the time of the robbery. The cops will have to cover the 65 miles in less than 30:22.5 minutes, or about 0.50625 hours.

The cops' speed then is 65 miles/(0.50625 hour) ≈ 128.39506... mi/h

he is right do not look down here to be sure so go with his answer

When the reference point is used in the point-slope form of the equation, the equation should evaluate to ...

... 0 = 0

This is only true for point (-16, 8) using the equation of selection a).

Happily, we also find that point (4, -2) satisfies the equation.

... -2-8 = -1/2(4+16)

... -10 = -1/2(20) . . . . true

The appropriate choice is ...

... a.) y-8 = -1/2(x+16)

Wording is everything. Here, there are some issues. "... at the rate of 1/2 per month" can be interpreted to mean that at the end of the first month, there are 649 1/2 items in Marie's closet (decreased  by 1/2 from 650).

"The number of items Dustin adds" could mean 5 items, the number he adds each month. The wording should specify the time period or whether we're talking about the total number Dustin has added.

We assume your description means that the number of items in Marie's closet at the end of each month is 1/2 what it was at the beginning. (As opposed to decreasing by 1/2 item each month.) We assume we're interested in the total number of items of Dustin's that are in the closet.

Marie's quantity can be modeled by ...

... m = 650·(1/2)^t . . . . . t = time in months

Dustin's quantity can be modeled by ...

... d = 5t

There will be one solution for d=m, at about t = 4.8. At that point, Dustin will have added about 24 items, which will be the number Marie is down to.

There is a viable solution for d=m at about t = 4.8.

Answer:

D. there is only one solution, and it is viable

Step-by-step explanation:

If all were general laborers, the payroll would be 93·34 = 3162 per day. The payroll is actually 3771 -3162 = 609 more than that.

Each equipment operator makes 29 more than a laborer, so there must be ...

... 609/29 = 21 heavy equipment operators

and 34 -21 = 13 general laborers

_____

If you write an equation for the number of heavy equipment operators (h), you get the same result in the same way. 34-h is the number of laborers

... 122h + 93(34 -h) = 3771 . . . equation for total daily payroll

... 29h + 3162 = 3771 . . . . . . simplify

... 29h = 609 . . . . . . . . . . . . . subtract 3162

... h = 609/29 = 21 . . . . . . . . divide by 29

The city has an initial population of 75,000.

Grows at a constant rate of 2,500 per year for five years

a) We must find a linear function that models the P population of the city according to the years.

This function has the following form:

[tex]P=P_{0}+ at\\[/tex]

Where

P is the population as a function of time

[tex]P_{0}[/tex] is the initial population

"a" is the constant rate of growth of the function.

"t" is the time elapsed in units of years.

Then the function is:

 [tex]P=75,000+2500t[/tex]

b) Before plotting the function, let's find its intercepts with the "t" and "P" axes

To find the intercept of the function with the t axis we do P = 0

 [tex]0 =75000+2500t[/tex]

 [tex]t=\frac{-75 000}{2500}[/tex]

 [tex]t = -30[/tex]

Now we make t = 0 to find the intercept with the P axis

[tex]P =75000[/tex]

The intercept with the P axis at P = 75 000 means that this is the initial population, therefore, for a period of 0 to 5 years, the population can not be less than 75,000.

The intercept at t = -30 does not have an important significance for this problem, since we are evaluating population growth for a period of [tex]0 \leq t \leq 5[/tex].

The graph of the function is shown in the attached figure.

c) To answer this question we must do P = 100 000 and clear t.

[tex]100000=75000+2500t [/tex]

[tex]25 000=2500t [/tex]

[tex]t =10years[/tex].

The second problem is solved in the following way:

The weight of a newborn baby is 7.5 pounds

The baby earns half a pound a month in its first year

a) To find the function that models the weight of the baby we follow the same procedure as in the previous problem.

[tex]W = W_{0} + at[/tex]

Where

W is the baby's weight according to the months

[tex]W_{0}[/tex] is the initial weight in pounds

"a" is the rate of increase

"t" is the time elapsed in months.

So:

[tex]W = 7.5 + 0.5t[/tex]

b) The domain of the function is [tex]0 \leq t \leq 12\\[/tex]

Since the function only applies for the first year of growth of the baby, and one year has 12 months.

The range of the function is [tex]7.5 \leq W\leq 13.5[/tex]

The towns' population and the baby's weight can be modeled by linear functions, which have a constant growth rate and an initial starting value. For the population to reach 100,000, we need to solve for t in our linear equation. Linear relationships are common in population growth, but are often approximations as they ignore limiting factors.

In both scenarios, we're dealing with linear functions. The towns' population, P, can be represented by a linear function as follows: P(t) = 2,500*t + 75,000 where t is the number of years passed. For the baby's weight, use the similar linear function: W(t) = 0.5*t + 7.5, where t is the baby's age in months. Both functions have an initial value (intercept at t=0) and a constant growth rate (the slope of the line). For the town's population to reach 100,000, solve the equation 100,000 = 2,500*t + 75,000 for t. Similarly, the baby's weight will depend on how many months have elapsed.

To graph either function, start at the intercept (t=0) and use the slope to find additional points (i.e., for each year that passes, add 2,500 to the population, or for each month that passes, add 0.5 to the baby's weight).

Linear relationships like these are common in Population Growth and other Population Models but are often approximates as they ignore factors that may limit growth (such as resources).

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If something goes below sea level, that means the answer is gunna be a negative number. so we have -200 and -130. we have to add the numbers together.

-200 + -130 = -330.

The new depth is -330. Hope this helps. Let me know if you need anymore help!

Answer:

(1, 3)

Step-by-step explanation:

The first number of each pair is the x-coordinate, the horizontal location of the point.

The end points of the graphed line segment are (1, 3) and (-3, -2). It looks like the answer choices are intended to see if you can tell what order the coordinates are expressed in.

The appropriate choice is (1, 3), the upper right end point.

[tex]\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{7})~\hspace{10em} slope = m\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=-2(x-4)[/tex]

Answer:

The equation in point-slope is [tex]y-7=-2(x-4)[/tex].

Step-by-step explanation:

Point-slope is a specific form of linear equations in two variables:

[tex]y-b=m(x-a)[/tex]

When an equation is written in this form, m gives the slope of the line and (a, b) is a point the line passes through.

We want to find the equation of the line that passes through (4, 7) and whose slope is -2. Well, we simply plug m = -2, a = 4, and b = 7 into point-slope form.

[tex]y-7=-2(x-4)[/tex]

19

evaluate the parenthesis, noting that + ( - ) = -

(19 + 9 ) + ( - 9) = 28 - 9 = 19

[tex]Solution, \left(19+9\right)+\left(-9\right)=19[/tex]

[tex]Steps:[/tex]

[tex]\mathrm{Follow\:the\:PEMDAS\:order\:of\:operations}[/tex]

[tex]\mathrm{Calculate\:within\:parentheses}\:\left(19+9\right)\::\quad 28, =28+\left(-9\right)[/tex]

[tex]\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:28+\left(-9\right)\::\quad 19, =19[/tex]

The correct answer is 19

Hope this helps!!!

y=x^2 - 10x +2

Use the form ax^2 + bx + c to find the values of a, b,and c.

a = 1 ( no number in front of the x^2)

b = 10

c = 2

Vertex form is a(x+d)^2 + e

Solve for d using d= b/2a

d = 10 / 2

d = 5

Find e using e = c - b^2/4a

e = 2 - 10^2/4

e = 2 - 25

e = -23

Vertex = d,e

Vertex = (5, -23)