All categories

3 years ago

Can I get help with this, please.

Mathematics

2 answers:

mamaluj [8]3 years ago

8 0

Answer:

Please find attached pdf

Step-by-step explanation:

Download pdf

NeTakaya3 years ago

6 0

Answer:

nnjncjnjnfjefnjefnesjnjfnsjnfsjes

Step-by-step explanation:

You might be interested in

Please help me!! ......

fenix001 [56]

Answer:

b)480 cm^2

Step-by-step explanation:

perimeter=(16+12+20) cm

=48 cm

lateral surface area= perimeter × height

=48×10 =480 cm^2

4 0

3 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 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

Four classmates collect data by conducting a poll. They

Stolb23 [73]

D. Aidan’s

The larger the sample size (number of people asked), the more accurate the results generally are, due to the law of large numbers.

4 0

2 years ago

If a circle has circumference c, what is it’s are in terms of C

Scilla [17]

$\text {Circumference = } 2 \pi r$

$c = 2 \pi r$

$\text {r = } \dfrac{c}{2 \pi }$

$\text {area = } \pi r^2$

Sub r into area:

$\text {area = } \pi ( \dfrac{c}{2 \pi } )^2$

$\text {area = } \dfrac{\pi c^2}{4 \pi^2 }$

$\text {area = } \dfrac{\ c^2}{4 \pi }$

7 0

2 years ago

The toatal charge on 6 is 48 units . what is the charge on each house

Vesna [10]

The charge on each house is 8 units

3 0

3 years ago