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ycow [4]
3 years ago
14

1. The number a is 4/5 of the number b. What part of number a is number b?

Mathematics
1 answer:
alexdok [17]3 years ago
7 0
Writing some equations out might help!

1) You know that a is 4/5 of b. This can be written as:a =  \frac{4}{5} b.
To find what b is equal to, multiply both sides by 5/4: \frac{5}{4}a = b.
A is 5/4 B. 

2) You know that a is 1/5 smaller than b. This can be written as: a = b - \frac{1}{5}.
To find b, just add 1/5 to both sides: a +  \frac{1}{5} = b.
B is 1/5 bigger than a. 
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5 0
3 years ago
Help ill give brainlest pls
Murrr4er [49]

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8 0
2 years ago
Vicky made a recipe for 144 fluid ounces of scented candle wax how many one cup candle molds can she fill with the recipe
san4es73 [151]
Well,

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3 0
3 years ago
Square roots in trigonometry. I don’t understand please help?
cupoosta [38]

By definitions of the (co)tangent and cosecant function,

3\tan^2x-2=\csc^2x-\cot^2x\iff3\dfrac{\sin^2x}{\cos^2x}-2=\dfrac1{\sin^2x}-\dfrac{\cos^2x}{\sin^2x}

Turn everything into fractions with common denominators:

\dfrac{3\sin^2x-2\cos^2x}{\cos^2x}=\dfrac{1-\cos^2x}{\sin^2x}

Recall that \cos^2x+\sin^2x=1, so we can simplify both sides a bit.

On the left:

\dfrac{3\sin^2x+3\cos^2x-5\cos^2x}{\cos^2x}=\dfrac{3-5\cos^2x}{\cos^2x}

On the right:

\dfrac{1-\cos^2x}{\sin^2x}=\dfrac{\sin^2x}{\sin^2x}=1

(as long as \sin x\neq 0, which happens in the interval 0\le x\le\pi when x=0 or x=\pi)

So we have

\dfrac{3-5\cos^2x}{\cos^2x}=1\implies3-5\cos^2x=\cos^2x

\implies3=6\cos^2x

\implies\cos^2x=\dfrac12

\implies\cos x=\pm\dfrac1{\sqrt2}

\implies x=\dfrac\pi4\text{ or }x=\dfrac{3\pi}4

4 0
3 years ago
What is The sum of b and 11
max2010maxim [7]
B+11 ? I’m not sure is this high school math ?
6 0
3 years ago
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