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2 years ago

I need help for part B could you show your work as well? thanks...

Mathematics

1 answer:

Ksivusya [100]2 years ago

8 0

Hello army :,) Here’s the answer:

B)
(3.19 * 10^8)/(8.23 * 10^5) = 3.19/8.23 * 10^8/10^5
= 3.19/8.23 * 10^3 = 0.387606318 * 10^3
= 387.606318

The population of US is 387 times larger than that of Ohio's

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The perimeter and the area of a rectangle have the same numerical value.

Sophie [7]

Formula

2*(L + W) = L * W

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W = 3

Solution

2*(L + 3) = 3*L

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2L - 2L + 6 = 3L - 2L combine

6 = L

Answer: The length is 6 units long.

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3 years ago

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3 years ago

(7 x 18 + 45) : 3 x 2 what is it please?

11Alexandr11 [23.1K]

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2 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

3 years ago

What is x if n≠m? Please help!

lorasvet [3.4K]

Answer: $x = \frac{mn(q - p)}{n-m}\\\\$

================================================

Work Shown:

Part 1

$\frac{x-m}{m} + p = \frac{x-n}{n} + q\\\\\frac{x-m}{m} - \frac{x-n}{n} = q - p\\\\\frac{n(x-m)}{mn} - \frac{m(x-n)}{mn} = q - p\\\\\frac{xn-mn}{mn} - \frac{xm-mn}{mn} = q - p\\\\\frac{xn-mn-(xm-mn)}{mn} = q - p\\\\$

Part 2

$\frac{xn-mn-xm+mn}{mn} = q - p\\\\\frac{xn-xm}{mn} = q - p\\\\\frac{x(n-m)}{mn} = q - p\\\\x(n-m) = mn(q - p)\\\\x = \frac{mn(q - p)}{n-m} \\\\$

-------------------------------

Explanations:

There are quite a bit of steps. I decided to break things into two parts.
The goal is to get x all by itself on its own side, which is why I subtracted (x-n)/n from both sides in part 1, step 2. I also subtracted p from both sides.
Afterward, I gave each fraction the LCD mn. I multiplied top and bottom of the first fraction by n/n. I did a similar operation to the second fraction, but with m instead.
From there, we distribute and simplify. The mn terms cancel on the left side numerator (second step of part 2).
The n≠m is there to prevent the denominator (n-m) from being zero. We cannot divide by zero.
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7 0

2 years ago