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3 years ago

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Draw the image of quadrilateral ABCD when translated by the directed line segment v. Label the image of A as A', the image of B

as B', the image of C as C' and the image of D as D'.

1 answer:

ICE Princess25 [194]3 years ago

3 0

Answer:

Try this on for size.

Step-by-step explanation:

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

Alla [95]

$\large \mathbb{PROBLEM:}$

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

$\large \mathbb{SOLUTION:}$

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

$\boxed{ \tt \red{C}arry \: \red{ O}n \: \red{L}earning} \: \underline{\tt{5/13/22}}$

4 0

2 years ago

There was a population of 80 fish in a National Park. After a year, the population increased by 5%. How many fish are there in t

Vaselesa [24]

Answer Its 84

Step-by-step explanation:

80 plus 5% equals 84%

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8 0

3 years ago

An irregular parallelogram rotates 360° about the midpoint of its diagonal. How many times does the image of the parallelogram c

Solnce55 [7]

The answer is 2 only

7 0

3 years ago

Use the law of wines to find the value of y. Round to the nearest Tenth

Eva8 [605]

Answer:

y ≈ 2.5

Step-by-step explanation:

Using the Sine rule in Δ XYZ

$\frac{sinZ}{XY}$ = $\frac{sinY}{XZ}$ , substitute values

$\frac{sin50}{2}$ = $\frac{sin75}{y}$ ( cross- multiply )

y sin50° = 2 sin75° ( divide both sides by sin50° )

y = $\frac{2sin75}{sin50}$ ≈ 2.5 ( to the nearest tenth )

4 0

3 years ago

Hey everyone. I need the answer to this ASAP!: Michael knows that 2 cans of paint is the exact amount he needs to paint a 10-foo

Ivahew [28]

Answer:

See attachment for graph

<em></em>

Step-by-step explanation:

Given

$2 cans = 10ft\ by\ 12ft$

Required

Graph the relationship between both parameters

First, calculate the area;

$Area = 10ft * 12ft$

$Area = 120ft^2$

This implies that: (2, 120) i.e. 2 cans for 120ft^2.

So we have:

(0,0) and (2,120)

(0,0) implies that: 0 cans for 0 square feet

Calculate the slope:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{120 - 0}{2-0}$

$m = \frac{120 }{2}$

$m = 60$

The equation is:

$y = m(x - x_1) + y_1$

So, we have:

$y = 60 * (x - 0) + 0$

$y = 60 * (x )$

$y = 60 x$

<em>See attachment for graph</em>

4 0

3 years ago