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3 years ago

Use substitution X+y=11 -x=-y-9

Mathematics

1 answer:

Alborosie3 years ago

4 0

Answer:

-(_y)-11=-y-9

Y-11=-y-9

2y=2

Y=1

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Can somebody please help 4=-0.8n

storchak [24]

4 = -0.8n

To find n divide both sides by -0.8:

n = 4 / -0.8

n = -5

5 0

3 years ago

Read 2 more answers

A cylindrical swimming pool has a diameter of 16 feet and a height of 3 feet. How many gallons of water can the pool contain? Ar

kiruha [24]

Answer:

The pool can contain 4521.6 gallons of water.

Step-by-step explanation:

Given:

The diameter of the cylindrical swimming pool = 16 feet.

And the height of the cylindrical swimming pool = 3 feet.

(1 ft³ = 7.5 gallons)

Now, to find the gallons of water the pool can contain.

So, we find the radius first:

Radius (r) = $\frac{Diameter}{2}$ = $\frac{16}{2} =8\ feet.$

Height (h) = 3 feet.

Now, to get the volume of swimming pool which is cylindrical in shape by putting formula (using the value of π = 3.14):

$Volume=\pi r^2h$

$Volume=3.14\times 8^2\times 3$

$Volume=3.14\times 64\times 3$

$Volume=602.88\ feet^3.$

Thus, getting the volume of pool 602.88 feet³.

As, given 1 ft³ = 7.5 gallons.

Now, to get the gallons of water the pool can contain we multiply the volume of pool by 7.5:

$602.88\times 7.5$

$=4521.6\ gallons.$

Therefore, the pool can contain 4521.6 gallons of water.

7 0

3 years ago

Suppose integral [4th root(1/cos^2x - 1)]/sin(2x) dx = A What is the value of the A^2?

Alla [95]

$\large \mathbb{PROBLEM:}$

$\begin{array}{l} \textsf{Suppose }\displaystyle \sf \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx = A \\ \\ \textsf{What is the value of }\sf A^2? \end{array}$

$\large \mathbb{SOLUTION:}$

$\!\!\small \begin{array}{l} \displaystyle \sf A = \int \dfrac{\sqrt[4]{\frac{1}{\cos^2 x} - 1}}{\sin 2x}\ dx \\ \\ \textsf{Simplifying} \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\sec^2 x - 1}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt[4]{\tan^2 x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sqrt{\tan x}}{\sin 2x}\cdot \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\tan x}{\sin 2x\ \sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{\sin x}{\cos x}}{2\sin x \cos x \sqrt{\tan x}}\ dx\:\:\because {\scriptsize \begin{cases}\:\sf \tan x = \frac{\sin x}{\cos x} \\ \: \sf \sin 2x = 2\sin x \cos x \end{cases}} \\ \\ \displaystyle \sf A = \int \dfrac{\dfrac{1}{\cos^2 x}}{2\sqrt{\tan x}}\ dx \\ \\ \displaystyle \sf A = \int \dfrac{\sec^2 x}{2\sqrt{\tan x}}\ dx, \quad\begin{aligned}\sf let\ u &=\sf \tan x \\ \sf du &=\sf \sec^2 x\ dx \end{aligned} \\ \\ \textsf{The integral becomes} \\ \\ \displaystyle \sf A = \dfrac{1}{2}\int \dfrac{du}{\sqrt{u}} \\ \\ \sf A= \dfrac{1}{2}\cdot \dfrac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \sqrt{u} + C \\ \\ \sf A = \sqrt{\tan x} + C\ or\ \sqrt{|\tan x|} + C\textsf{ for restricted} \\ \qquad\qquad\qquad\qquad\qquad\qquad\quad \textsf{values of x} \\ \\ \therefore \boxed{\sf A^2 = (\sqrt{|\tan x|} + c)^2} \end{array}$