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3 years ago

5s-(100+2s =4s-2(10+s) how much boxes would they need to sell to get a equal amount each group must sell to have equal profits.

Mathematics

1 answer:

antoniya [11.8K]3 years ago

6 0

5s-100-2s=4s-20-2s
3s-100. = 2s-20
+20. +20
3s-80. = 2s
-3s. -3s
-80. = -1s
-1 -1
80. = s

s=80

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-3(7w-3)+8w=-5(w+1) what is the value of w

bogdanovich [222]

Answer:

Step-by-step explanation:

Step 1:

- 21w + 9 + 8w = - 5w - 5

Step 2:

- 13w + 9 = - 5w - 5

Step 3:

- 7w + 9 = - 5

Step 4:

- 7w = - 14

Step 5:

14 = 7w

Answer:

w = 2

Hope This Helps :)

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3 years ago

If the parent function is y=4*, which is the function of<br> the graph?

Vaselesa [24]

Answer:

The answer is y = 3(4)^x or answer choice C

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2 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to <em>x</em> :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to <em>y</em> :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

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3 years ago

Find the surface area and volume of each figure. Round each answer to the nearest hundredth.

Llana [10]

Answer: Surface area is equal to 200 $cm^{2}$

Volume is equal to 333.33 $cm^{3}$

Step-by-step explanation:

First, let's do surface area.

The surface area of a pyramid is equal to 1/2(perimeter of base)(lateral height) + area of the base

The perimeter of the base is 10(4) = 40; as the base is a square with a side length of 10.

The lateral height is given as 5 cm.

The area of the base is 10(10) = 100.

We can plug those numbers into the equation to get 1/2(40)(5) + 100, which comes out to be 200 $cm^{2}$ .

Now for volume.

The volume of a pyramid is equal to 1/3(area of the base)(height).

We already have the area of the base, which is 100.

The height is given as 10 cm.

Plugging those numbers into the equation, we get 1/3(100)(10), which is 1000/3 or about 333.33 $cm^{3}$ .

Hope this helps!

6 0

3 years ago

Which expression has a value that is equivalent to the given expression when n = 8?

mrs_skeptik [129]

Q1.

Put the value of n = 8 to the expressions:

2n + 10 → 2(8) + 10 = 16 + 10 = 26

A: 3n → 3(8) = 24 ≠ 26

B: 2(n + 5) → 2(8 + 5) = 2(13) = 26 :)

C: n² + 5 → 8² + 5 = 64 + 5 = 69 ≠ 26

D: 10n + 2 → 10(8) + 2 = 80 + 2 = 82 ≠ 26

<h3>Answer: B: 2(n + 5)</h3>

Q2.

x = 3, y = 2

L = 5y² - 2x → L = 5(2)² - 2(3) = 5(4) - 6 = 20 - 6 = 14

R = x² + ... → R = 3² + ... = 9 + ...

A: x² + x → 9 + 3 = 12 NOT

B: x² + y → 9 + 2 = 11 NOT

C: xy → 9 + (3)(2) = 9 + 6 = 15 YES

D: 2y → 9 + (2)(2) = 9 + 4 = 13 NOT

5y² - 2x < x² + xy

L = 14, R = 15, L < R CORRECT

<h3>Answer: C: xy</h3>

Q3.

3 + 6² = 3 + 36 = 39

A: 3³ + 12 = 27 + 12 = 39 NOT

B: 5² + 4² = 25 + 16 = 41 > 39 YES

C: 2³ + 5² + 6 = 8 + 25 + 6 = 39 NOT

D: 2³ + 14 + 17 = 8 + 14 + 17 = 39 NOT

<h3>Answer: B: 3 + 6² < 5² + 4²</h3>

6 0

3 years ago