3 years ago

4 5/7-1 6/7 in simplest form please show your work or explain thank you

Mathematics

1 answer:

RoseWind [281]3 years ago

7 0

Put both fractions into an improper fraction.... 33/7 - 13/7. then you subtract the numerators.. so its 20/7 or 2 6/7.

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2 years ago

3 square root 125/5-8

Ksivusya [100]

The answer is -1.2917960675

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}

VashaNatasha [74]

Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

$\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}$

Rationalize the denominator by multiplying both numerator and denominator by √10:

$\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}$

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

$\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}$

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2 years ago

For the right triangle shown, the lengths of two sides are given. find the third side. leave your answer in simplified, radical

lisov135 [29]

If c is the hypotenuse and b is the leg, then the third side (a):
$a= \sqrt{c^2-b^2}= \sqrt{16^2-9^2}= \sqrt{256-81}= \sqrt{175}=5 \sqrt{7}$

If b and с are legs, then hypotenuse (a):
$a= \sqrt{b^2+c^2}= \sqrt{9^2+16^2}= \sqrt{81+256}= \sqrt{337}$

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2 years ago

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Form the union for the following sets. R = {10, 15, 20} S = {20, 25} R ∪ S =

Alja [10]

R = {10; 15; 20}

S = {20; 25}
R U S = {10;15; 20; 25;

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3 years ago