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3 years ago

Can someone please help me solve this math problem? Thank you!

Mathematics

1 answer:

sattari [20]3 years ago

3 0

A; x= 1/2
y= 6/12
z= 50/100

B; 50%

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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

Would each of them be able to be the side lengths of a triangle?

ratelena [41]

Pythagoras Theorem:
hipotenuse²=leg₁²+leg₂²

First posible triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=5 (5²=25)

13³=144 + 25

Answer:can be side lengths of a triangle

Second triangle:
hypotenuse=12.6 (12.6²=158.76)
leg₁=6.7 ( 6.7²=44.89)
leg₂=6.5 (6.5²=42.25)

leg₁²+leg₂²=44.89+42.25=87.14≠158.76

Answer: cannot be side lenghts of a triangle.

third triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=11 (11²=121)

leg₁²+leg₂²=144+121=265≠169

Answer: cannot be side lenghts of a triangle.

fourth triangle:
hypotenuse=13 (13²=169)
leg₁=6 ( 6²=36)
leg₂=4 (4²=16)

leg₁²+leg₁²=36+16=52≠169

Answer: cannot be side lenghts of a triangle.

8 0

3 years ago

Read 2 more answers

What is bigger 9/10 or 7/8

Nataliya [291]

You start by cross multiplying:
9 x 8 = 72
This is considered the fraction on the left's value.
You do the same for the other side:
10 x 7 = 70
This is considered the value of the fraction 7/8
72 has a greater value than 70.

Therefore, 9/10 is larger.

4 0

4 years ago

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3) If you can buy one package of shallots for<br> $4, then how many can you buy with $12?

Olin [163]

Answer:

Step-by-step explanation:

4$ = 1 pack

(think: what times 4 is 12? 3! so we need to muliply both sides of the equal sign by 3, so we can turn the 4 into a 12. Remember, what you do to on side, you must do to the other. )

4$ = 1 pack

*3 *3

12$ = 3 packs

so your answer is 3.

5 0

3 years ago

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Which number had a 3 with a value that is 100 times greater than the value of the 3 in 20.342?

goblinko [34]

Here the value of 3 is =3/10
Then 100 fold of 3/10=3/10*100=30
So, the number 26538,532,97436 etc. has that property.

4 0

3 years ago