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3 years ago

32÷891 using long division

Mathematics

2 answers:

valentinak56 [21]3 years ago

5 0

It would be 0.03591470258136924803591470258136924803

Alinara [238K]3 years ago

5 0

It would be 27 with a remainder

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What a ratio of 4 days to 36 hours

klemol [59]

Answer:

4:36

24 x 4:36

96:36

8:4

1day = 24 hours

Step-by-step explanation:

7 0

3 years ago

Can someone please help me with this?

Andru [333]

Answer:

Option D

Step-by-step explanation:

Convert to slope-intercept form:

$4x-2y=2$

$-2y=2-4x$

$y=-1+2x$

$y=2x-1$

Find the x-intercept by plugging in y=0:

$y=2x-1$

$0=2x-1$

$1=2x$

$\frac{1}{2}=x$

$x=\frac{1}{2}$

Therefore, the x-intercept is $(\frac{1}{2},0)$

Find the y-intercept by plugging in x=0:

$y=2x-1$

$y=2(0)-1$

$y=0-1$

$y=-1$

Therefore, the y-intercept is $(0,-1)$

So, the correct answer is option D.

4 0

2 years ago

Read 2 more answers

Find the area of the rectangle

alexira [117]

112ft squared since to find the area of a rectangle you have to do length times width and 15*8 gives you 112

4 0

4 years ago

Help will give thanks to the first to answer

satela [25.4K]

Answer:

$56$

Step-by-step explanation:

$10 \times 5 + 6$

$50 + 6 = 56$

7 0

2 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