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2 years ago

How to convert improper fraction to proper fraction

1 answer:

6 0

Multiply the whole number by the denominator. Add the product to the numerator of the proper fraction. The sum is the numerator of the improper fraction. The denominator will stay the same.

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Solve for the variable r in this equation: s = 2_rh

svp [43]

I hope this helps you

s=2.r.h

r=s/2h

3 0

3 years ago

Read 2 more answers

I need helps can you simply this please :)

Gwar [14]

Answer:

64-18+2

Hope this helps...

6 0

2 years ago

Someone please help me! Studying for finals and don’t understand how to complete problems like this! Thank you!

Drupady [299]

Answer:

(d). $\frac{152}{377}$

Step-by-step explanation:

sin² β + cos² β = 1

cos (α + β) = cos α cos β - sin α sin β

cos α = √(1 - $\frac{400}{841}$ ) = 21 / 29

cos β = 12 / 13

sin β = √(1 - $\frac{144}{169}$ ) = 5 / 13

sin α = 20 / 29

cos (α + β) = $\frac{21}{29}$ × $\frac{12}{13}$ - $\frac{20}{29}$ × $\frac{5}{13}$ = $\frac{152}{377}$

6 0

3 years ago

Andres is offered a job as manager of a health club. The job will pay $3,100 per month, paid

dybincka [34]

The answer is C hope it helps

7 0

2 years ago

Lim x-0 (sin2xcsc3xsec2x)/x²cot²4x

sergij07 [2.7K]

By the definitions of cosecant, secant, and cotangent, we have

$\dfrac{\sin2x\csc3x\sec2x}{x^2\cot^24x}=\dfrac{\sin2x\sin^24x}{x^2\sin3x\cos2x\cos^24x}$

Then we rewrite the fraction as

$\dfrac{\sin2x}{2x}\left(\dfrac{\sin4x}{4x}\right)^2\dfrac{3x}{\sin3x}\dfrac{32}{3\cos2x\cos^24x}$

The reason for this is that we want to apply the well-known limit,

$\displaystyle\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1$

for $a\neq0$ . So when we take the limit, we have

$\displaystyle\lim_{x\to0}\cdots=\lim_{x\to0}\frac{\sin2x}{2x}\left(\lim_{x\to0}\frac{\sin4x}{4x}\right)^2\lim_{x\to0}\frac{3x}{\sin3x}\lim_{x\to0}\frac{32}3\cos2x\cos^24x}$

$=1\cdot1^2\cdot1\cdot\dfrac{32}3=\dfrac{32}3$

8 0

3 years ago