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9 months ago

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Triangle abc has side lengths ab = 65, bc = 33, and ac = 56. find the radius of the circle tangent to side ac and bc and to the

circumcircle of triangle abc.

1 answer:

lara [203]9 months ago

7 0

The radius of the circle tangent to sides AC and BC and to the circumcircle of triangle ABC.

r= 24.

<h3>What is the radius of the circle tangent to sides AC and BC and to the circumcircle of triangle ABC.?</h3>

Generally, the equation for side lengths AB is mathematically given as

Triangle ABC has side lengths

Where

AB = 65,
BC = 33,
AC = 56.

Hence

r √ 2 · (89 √ 2/2 − r √ 2) = r(89 − 2r),

r = 89 − 65

r= 24.

In conclusion, The radius of the circle tangent to sides AC and BC and to the circumcircle of triangle ABC.

r= 24.

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Given that the expression 2x^3 + mx^2 + nx + c leaves the same remainder when divided by x -2 or by x+1 I prove that m+n =-6

Alla [95]

Given:

The expression is:

$2x^3+mx^2+nx+c$

It leaves the same remainder when divided by x -2 or by x+1.

To prove:

$m+n=-6$

Solution:

Remainder theorem: If a polynomial P(x) is divided by (x-c), thent he remainder is P(c).

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$P(x)=2x^3+mx^2+nx+c$

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$P(-1)=-2+m-n+c$

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$P(2)=16+4m+2n+c$

Now, substitute the values of P(2) and P(-1) in (i), we get

$16+4m+2n+c=-2+m-n+c$

$16+4m+2n+c+2-m+n-c=0$

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$3m+3n=-18$

Divide both sides by 3.

$\dfrac{3m+3n}{3}=\dfrac{-18}{3}$

$m+n=-6$

Hence proved.

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