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kkurt [141]
3 years ago
11

GEOMETRY. NEED HELP NOW PLEASE!!

Mathematics
2 answers:
yan [13]3 years ago
4 0
So here we go, lines OA and OB are radii of the circle O
since the measurement of ∡ O is given it is easy to find the other measurements.
180-85 = 95°  (all the sides of a triangle = 180°)
That angle divided by two
\frac{95}{2} = 47.5°
We know that the radii are the sides which makes that triangle isosceles.
So m∡BAC = 47.5°
Answer is B)47.5
11111nata11111 [884]3 years ago
4 0
They have already given you d. 360 divided by 2 is your answer
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Verify that -5, 1/2, and 3/4 are the zeroes of the cubic polynomial 4x^3+20x^2+2x-3. Also verify the relationship between the ze
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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\\

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Hence we do not find any relation between the coefficients and zeros.

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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