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2 years ago

Ind the approximate increase in its area if its radius increases from mm to mm. [Recall that the area of a circle is A.]

Mathematics

1 answer:

nlexa [21]2 years ago

7 0

Answer:

Increase in area of circle is $263.76$ square mm

Step-by-step explanation:

The actual question is

Find the approximate increase in its area if the radius increases from 20 mm to 22 mm.

Solution

The area of a circle of radius r is given by

$A = \pi r^2$

Area of circle of radius 20 mm is

$A_1 = \pi 20^2\\A_1 = 400\pi$ square mm

Area of circle of radius 22 mm is

$A_2 = \pi 22^2\\A_2 = 484\pi$ square mm

Increase in area of circle

$A_2 - A_1 = 484 \pi - 400 \pi \\A_2 - A_1 = 84 \pi \\A_2 - A_1 = 263.76$ square mm

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(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 µ(x, y) 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 µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

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

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 F(x, y) = C. 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 x :

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

Differentiate both sides of this with respect to y :

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

Then the general solution to the DE is

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

5 0

3 years ago

Someone Please help me answer this question.

Shalnov [3]

∠C is 148°.

Step-by-step explanation:

Step 1: Given ∠A = 16° and ∠B = (10x - 64)°. Also, AC = BC. Now, the angles opposite to equal sides of a triangle are equal. So ∠B = 16°
Step 2: Sum of angles of a triangle is 180°

⇒ 16° + 16° + ∠C = 180°

⇒ ∠C = 180 - 32 = 148°

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3 years ago

Quick Fix Inc. repairs bikes. Their revenue, in dollars, can be modeled by the equation y = 400 + 220x, where x is the number of

Sholpan [36]

The break even point is that where the total cost is equal to the total revenue. To determine the number of hours in which Quick Fix Inc. will have a breakeven, we have to equate the two equations and solve for the value of x,
400 + 220 x = 20x² + 160
The values of x from the equation are 12 and -1. Since, we do not have a negative for the number of hours, our answer is 12 hours.

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