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natka813 [3]
3 years ago
10

Which of the following ratios is equivalent to 5/8 ?

Mathematics
2 answers:
Novosadov [1.4K]3 years ago
8 0

Answer:

Which of the following ratios is equivalent to 5/8 ?

A) 16 : 10

B) 5 to 13  

C) 25/44

<u>D) 15 out of 24</u>

Step-by-step explanation:

5 : 8

(×3)

15 : 24

svetoff [14.1K]3 years ago
3 0

The answer is D.....

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Take the expression in chunks:

>>

((X+A)(X-A)M+A^2M)E=((X^2-A^2)M+A^2M)E=X^2ME

>>

\dfrac{\frac{R+X}X+\frac X{R-X}}{\frac X{R-X}}=\dfrac{\frac{R+X}X}{\frac X{R-X}}+1

\dfrac{\frac{(R+X)(R-X)}{X(R-X)}}{\frac{X^2}{X(R-X)}}=\dfrac{(R+X)(R-X)}{X^2}=\dfrac{R^2-X^2}{X^2}=\dfrac{R^2}{X^2}-1

\implies\dfrac{\frac{R+X}X+\frac X{R-X}}{\frac X{R-X}}=\dfrac{R^2}{X^2}

>>

Note that (a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab. So

\displaystyle\frac{4AS}3\sum_{\rm cyc}\frac1{X((E+Y)^2-(E-Y)^2)}=\frac{4AS}3\sum_{\rm cyc}\frac1{4XEY}=\frac{AS}3\sum_{\rm cyc}\frac1{XEY}

where the cyclic sum notation means

\displaystyle\sum_{\rm cyc}f(x,y,z)=f(x,y,z)+f(z,x,y)+f (y,z,x)

In other words, we take the sum over all possible cycles of the sequence of arguments to the summand. The second square-bracketed chunk reduces to

\displaystyle\frac{AS}3\sum_{\rm cyc}\frac1{XEY}=\frac{AS}3\left(\dfrac1{XEY}+\dfrac1{EYX}+\dfrac1{YXE}\right)=\dfrac{AS}3\dfrac3{XEY}=\dfrac{AS}{XEY}

So to recap, we've reduced the starting expression to

X^2ME\left(\dfrac{R^2}{X^2}-\dfrac{AS}{XEY}\right)Y

>>

\dfrac{R^2}{X^2}-\dfrac{AS}{XEY}=\dfrac{R^2EY-ASX}{X^2EY}

Finally, we have

X^2ME\dfrac{R^2EY-ASX}{X^2EY}Y=M(R^2EY-ASX)

Then distributing M and rewriting to decode the message, we have

MERRY-XMAS

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