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sweet-ann [11.9K]
3 years ago
7

Cannot solve this question need help asap

Mathematics
1 answer:
Schach [20]3 years ago
3 0

P(0,0) and R(2b, 2c)

Midpoint:

x = (x1+x2)/2 and y=(y1+y2)/2

Midpoint PR

x = (0 + 2b)/2

= 2b/2

=b

y = (0 + 2c)/2

=2c/2

=c

Midpoint PR (b,c)

Answer is the last one

(b,c)

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<em>All </em><em>real </em><em>numbers </em><em>are </em><em>greater </em><em>than </em><em>four</em>

Step-by-step explanation:

4+2=6

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Multiplying polynomials <br> (4x+3y)(5x+y)
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F(x) = X5 + 5x4 + 20x3 - 96x - 80 Has Some Of Its Zeros As X = -2, -1,2 What Would Be The Factorization Of F(x)? A. (x + 2)(x +
Svetlanka [38]

Answer:

\left(x+1\right)\left(x-2\right)\left(x+2\right)(x +2 - 4i)(x +2 + 4i)

Step-by-step explanation:

Given

f(x) = x^5 + 5x^4 + 20x^3 - 96x - 80

Zeros: -2, -1, 2

Required

Factorization of f(x)

The given zeros imply that:

x=-2     x =-1    x = 2

This gives:

x + 2 =0        x + 1 = 0       x - 2 = 0

So, some factors are:

(x + 2), (x + 1) and (x - 2)

Divide f(x) by the factors, to get the other factors:

\frac{x^5 + 5x^4 + 20x^3 - 96x - 80}{(x + 2)(x + 1) (x - 2)}

Using a factorization calculator, we have:

\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x^2+4x+20\right)}{(x + 2)(x + 1) (x - 2)}

Cancel out common terms

x^2+4x+20

Using quadratic formula, we have:

x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}

Where

a = 1; b =4; c = 20

x = \frac{-4 \± \sqrt{4^2 - 4*1*20}}{2*1}

x = \frac{-4 \± \sqrt{16 - 80}}{2*1}

x = \frac{-4 \± \sqrt{-64}}{2}

Using complex notation

\sqrt{-64}= 8i

So:

x = \frac{-4 \± 8i}{2}

Simplify the fraction

x = -2 \± 4i

Split

x = -2 + 4i \ or\  x = -2 - 4i

Equate to 0

x +2 - 4i = 0 \ or\ x +2 + 4i = 0

The other factors are: (x +2 - 4i) and (x +2 + 4i)

Hence, the factorization of f(x) is:

\left(x+1\right)\left(x-2\right)\left(x+2\right)(x +2 - 4i)(x +2 + 4i)

7 0
3 years ago
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