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Pachacha [2.7K]
2 years ago
10

If function g has the factors (x - 7) and (x + 6), what are the zeros of function g?

Mathematics
2 answers:
vagabundo [1.1K]2 years ago
8 0

The zeroes of the function will be equal to ( 7, -6 ).

<h3>What is a quadratic equation?</h3>

It is a polynomial with a degree of 2 or the maximum power of the variable is 2 in quadratic equations. It has two solutions as its maximum power is 2.

Given function:-

  • ( x- 7 ) ( x+ 6 )

As we can see that the quadratic equation in the factorized form is ( x - 7 ) ( x+ 6 ) so we will equate each factor to zero.

( x- 7 )  = 0

x = 7

( x+ 6 ) = 0

x = -6

Therefore zeroes of the function will be equal to ( 7, -6 ).

To know more about quadratic equations follow

brainly.com/question/1214333

#SPJ1

vaieri [72.5K]2 years ago
7 0
<h2>Function Equations</h2>

When a given function is written with factors in the form of ( x - r ), then <em>r</em> is a zero of the function.

<h2>Solving the Question</h2>

We're given:

  • Function <em>g</em> has factors (x-7) and (x+6)

Therefore, the zeros of the function are 7 and -6.

<h2>Answer</h2>

-6, 7

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~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-26}~,~\stackrel{y_2}{120})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{(~~-26 - 0~~)^2 + (~~120 - 0~~)^2} \implies r=\sqrt{(-26)^2 + (120 )^2} \\\\\\ r=\sqrt{( -26 )^2 + ( 120 )^2} \implies r=\sqrt{ 676 + 14400 } \implies r=\sqrt{ 15076 } \\\\[-0.35em] ~\dotfill

\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-26}{h}~~,~~\underset{120}{k})}\qquad \stackrel{radius}{\underset{\sqrt{15076}}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-26) ~~ )^2 ~~ + ~~ ( ~~ y-120 ~~ )^2~~ = ~~(\sqrt{15076})^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x+26)^2+(y-120)^2 = 15076~\hfill

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1 year ago
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