All categories

3 years ago

Please help me with my math question.

Mathematics

1 answer:

stealth61 [152]3 years ago

4 0

please subscribeInformal letter

1.since April 2020, you haven't gone outside ,not visited any place you isolated yourself for safety . Write a letter to your friend describing him oblic her about how you spent that time in home , how became meantaly strong.200-250

You might be interested in

Keb wants to put a swimming pool in his backyard. He buys a square tarp to go under it. The circumference of the pool is 48 feet

creativ13 [48]

Answer: bro i do not know the answer but you should ask someone different.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the measure of the indicated angle

emmasim [6.3K]

I think the answer is 23

8 0

3 years ago

Read 2 more answers

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

Alla [95]

$\large \mathbb{PROBLEM:}$

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

$\large \mathbb{SOLUTION:}$

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

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

4 0

2 years ago

Write the equation in point-slope form of the line that passes through the given point and has the given slope (4,-7); m= -1/4

kicyunya [14]

Answer:

y + 7 = -1/4(x - 4)

General Formulas and Concepts:

Algebra I

Point-Slope Form: y - y₁ = m(x - x₁)

x₁ - x coordinate
y₁ - y coordinate
m - slope

Step-by-step explanation:

Step 1: Define

Slope m = -1/4

Point (4, -7)

Step 2: Write Function

y + 7 = -1/4(x - 4)

6 0

3 years ago

I'll give brainliest if u are right

emmainna [20.7K]

Answer:

0.3

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers