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Elden [556K]
3 years ago
5

8/11(n-10)=64 I need to show my work to

Mathematics
2 answers:
Bogdan [553]3 years ago
6 0
Hope you understand it and see what I did.

Volgvan3 years ago
3 0
So first you multiply 8 over 11 and n and you get: 8/11n
then you multiply 8 over 11 and 10 and you get 80 over 11 but then you have to make it a mixed fraction which is 7 1 over 2. then you end up with this equation: 8 over 11n X 7 1 over 2 equals 64. Then im stuck.. because im in 8th grade and in Pre Algebra so im just now learning this.. xD but maybe you can get it from there.
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HELPP WILL MARK BRAINLIEST<br><br> 0.167 ml =blank l
likoan [24]

There are 1000 ml in 1 liter so

<u>1000ml</u> = <u>0.167ml</u>

  1liter           x

cross multiply to get 1000x = 0.167; divide both sides by 1000 to get 0.000167 liters

6 0
4 years ago
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
EFGH is a parallelogram. Find the measure of the following.
Sergeu [11.5K]
W=4
JG=12
This is the answer
4 0
3 years ago
Which of the following expressions represents the GCF of 91x 2 y and 104xy 3?
zlopas [31]
The greatest common factor is found by finding the product of common primes.

91=7*13, 104=2*2*2*13  so the gcf of 91 and 104 is 13.  Since the highest power of x and y in both terms is 1, the hcf for the variables is just xy

13xy
6 0
3 years ago
Read 2 more answers
A vet weighs two puppies. The small puppy weighs 4 2/3 pounds. The large puppy weighs 4 2/3 times as much as the small puppy. Ho
Maksim231197 [3]
4 2/3 lbs = small puppy weight
4 2/3 times the small puppies weight = large puppy weight
          4 2/3 × 4 2/3  or 4 2/3^2  = 21.77777.... or in mixed number is 21 7/9 lbs
4 0
4 years ago
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