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3 years ago

Help, teachers asking me to do this independently but I barely understand. Help ASAP!!!

Mathematics

1 answer:

Iteru [2.4K]3 years ago

4 0

Answer:

Step-by-step explanation:

2 x 2 x 2 = 8

8 - (8 - 4) + 5

(8 - 4)= 4

8 - 4 = 4

4 + 5 = 9

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Erin decided to spend $650 to install carpet in her square bedroom. Each side is 22feet long.

Bas_tet [7]

22 x 4 = 88 feet(c) or 650 X 2 = 1300

6 0

3 years ago

Cullumber company had $170,000 of net income in 2021 when the selling price per unit was $152, the variable costs per unit were

taurus [48]

The number of units sold in 2021 = 13,300 units.

The number of units that would have to be sold in 2022 to reach the stockholder's desired profit level = 15,200 units.

Assuming that Cullumber company sells the same number of units in 2022 as it did in 2021, the selling price in order to reach the stockholder's desired profit level = $160.

The net income of a company is its profit function, derived by the formula:

Profit = Sales - Cost.

Sales are the product of the number of units sold and per unit selling price.

Cost is the sum of fixed costs and the product of the number of units produced to the variable cost per unit.

Assuming the number of units sold in 2021 to be x,

we can say sales = $152x, and

cost = ${574800 + 96x}.

Therefore, profit = sales - cost,

or, 170000 = 152x - {574800 + 96x},

or, 170000 = 56x - 574800

or, 56x = 574800 + 170000 = 744800,

or, x = 744800/56 = 13300.

Therefore, the number of units sold in 2021 was 13,300.

In 2022, the desired profit = $170,000 + $106,400 = $276,400.

Assuming the number of units sold in 2022 to be n, at the same unit data and fixed cost, we can say that:

Sales = $152n, and

Cost = ${574800 + 96n}.

Therefore, profit sales - cost,

or, 276400 = 152n - {574800 + 96n},

or, 276400 = 56n - 574800,

or, 56n = 574800 + 276400 = 851200,

or, n = 851200/56 = 15200.

Therefore, the number of units sold in 2022 should be 15,200 to reach the desired profit of the stakeholders.

When the number of units sold in 2022 is the same as in 2021, that is, the number of units = 13,300, we assume the per unit selling price to be $P.

Now we can say that,

Sales = $13300P.

Costs = ${574800 + 96*13300} = ${574800 + 1276800} = $1851600.

Therefore, profit = sales - cost,

or, 276400 = 13300P - 1851600,

or, 13300P = 1851600 + 276400 = 2128000,

or, P = 2128000/13300 = 160.

Therefore, the new selling price per unit to reach stockholders' desired profit, assuming the number of units sold in 2022 to be the same as in 2021 is $160.

Refer to the attachment for the required questions.

Learn more about selling price, cost price, and profits at

brainly.com/question/19104371

#SPJ4

5 0

2 years ago

What is the answer for 6x+3y=15 in a order paired

melomori [17]

Hi there, the answers for this would be:

(0 , 5 ) , ( 1 , 3 ) , ( 2 , 1 )

Step-by-step explanation:

The problem is looking for graph coordinates that correspond to the given equation.

6x+3y=15

To get the same results as I did you would need to substitute the x,y for a set of numbers that would result in the answer to the equation which is 15.

6*0+3*5=15

6*1+3*3=15

6*2+3*1=15

Hope this helps and I hope you have a great day!

God Bless,

~ASIAX

4 0

3 years ago

Solve for n 5 5/6 +n = 10 1/12

Kobotan [32]

Answer:

Exact form:

N = 17/4

Decimal form:

N = 4.25

Mixed Number form:

N = 4 1/4

Step-by-step explanation:

Solve for N by simplifying both sides of the equation, then isolating the variable.

4 0

3 years ago

(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^

Sedbober [7]

Hello,

a)
$I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\

= [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\

=0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\

= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\
= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\

$
$I(1+n-1)= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
I= \dfrac{n-1}{n} *\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
$

b)
$\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\
= \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\
= \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\




$

$\int\limits^{ \frac{\pi}{2} }_0 {sin^{5}(x)} \, dx \\
= \dfrac{4}{5}*\dfrac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx = \dfrac{8}{15}\\





$

c)

$I_n= \dfrac{n-1}{n} * I_{n-2} \\

I_{2n+1}= \dfrac{2n+1-1}{2n+1} * I_{2n+1-2} \\
= \dfrac{2n}{2n+1} * I_{2n-1} \\
= \dfrac{(2n)*(2n-2)}{(2n+1)(2n-1)} * I_{2n-3} \\
= \dfrac{(2n)*(2n-2)*...*2}{(2n+1)(2n-1)*...*3} * I_{1} \\\\

I_1=1\\

$

3 0

4 years ago