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Mathematics

1 answer:

DanielleElmas [232]3 years ago

8 0

Solution is in the attachment.

(-3, 5, 6)

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Rewrite (2x+1)(x-2) using the distributive property then simplify

Contact [7]

(2x+1)(x-2) = 2x^2 - 4x +x -2= 2x^2-3x-2

3 0

4 years ago

Help? I'm not sure I understand how to solve this.

KiRa [710]

Okay so CBE is a right angle which equals 90 degrees. CBD and DBE makes up CBE which is 23 degrees is part of CBD. So subtract CBE from CBD to get DBE which is 67 degrees.

8 0

4 years ago

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$