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Alekssandra [29.7K]
2 years ago
6

PLEASE ANSWERRR I WILL GIVE TEN POINTS ITS EASY I SWEAR IM JUST NOT CAPABLE BECAUSE OF MY OTHER ASSIGNMENT PLES IM BEGGING

Mathematics
1 answer:
kvv77 [185]2 years ago
4 0

1) In order to add or subtract fractions denominator must be equal

\frac{1}{4}  +  \frac{2}{5}

Lcm of denominator is 4×5=20 since they are co primes

\frac{1}{4}  \times  \frac{5}{5}  +  \frac{2}{5}  \times  \frac{4}{4}

Now the denominator becomes same

\frac{5 + 8}{20 }  =  \frac{13}{20}

2)

\frac{3}{8}  -  \frac{1}{2}

Lcm =8

\frac{3}{8}  \times  \frac{1}{1}   -  \frac{1}{2}  \times  \frac{4}{4}

\frac{3 - 4}{8}  =  \frac{ - 1}{8}

3)

We just need to add the fractions

\frac{1}{4}  +  \frac{1}{3}

Lcm=12

\frac{1}{4}  \times  \frac{3}{3}  +  \frac{1}{3}  \times  \frac{4}{4}

\frac{3 + 4}{12}  =  \frac{7}{12}

hope it helps :)

plz mark as brainliest

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Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

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$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

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