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jasenka [17]
3 years ago
6

What is the square root of 25/49

Mathematics
2 answers:
kolbaska11 [484]3 years ago
6 0
First take the square root of 25, then take the square root of 49.
It is as easy as separating the square root to apply to both terms separately instead of combined.

sqrt(25)/sqrt(49)
5/7
makvit [3.9K]3 years ago
5 0
(25/49)^1/2= 5/7 where 25^1/2=5 and 49^1/2 is equal to 7
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Solution:

we have been asked to verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial 4x^3+20x^2+2x-3

To verify that whether the given values are zeros or not we will substitute the values in the given Polynomial, if it will returns zero, it mean that value is Zero of the polynomial. But if it return any thing other than zeros it mean that value is not the zero of the polynomial.

Let f(x)=4x^3+20x^2+2x-3\\\\\text{when x=-5}\\\\f(-5)=4(-5)^3+20(-5)^2+2(-5)-3=-13\\\\\text{when x=}\frac{1}{2}\\

f( \frac{1}{2} ) = 4 ( \frac{1}{2} )^3+20(\frac{1}{2})^2+2(\frac{1}{2})-3=\frac{7}{2}\\\\

\text{when x=}\frac{3}{4}\\\\

f( \frac{3}{4} ) = 4 ( \frac{3}{4} )^3+20(\frac{3}{4})^2+2(\frac{3}{4})-3=\frac{183}{16}\\

Hence -5, 1/2, and 3/4 are not the zeroes of the given Polynomial.

Since sum of roots=\frac{-b}{a}= \frac{-20}{4}=-5\\

But -5+\frac{1}{2}+\frac{3}{4}=  \frac{-15}{4}\neq-5

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Anyway if the given values doesn't represents the zeros then those given values will not have any relation with the coefficients of the p[polynomial.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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