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Andrew [12]
3 years ago
8

15 Points!

Mathematics
2 answers:
Nezavi [6.7K]3 years ago
8 0
Replace f(x) with 0 and swith the equation around to solve for x

 x^4 - 32x^2 - 144 = 0

Factor the left side:

(x+6) (x-6) (x^2+4)

so far we have x+6 = 0, x = -6
                         x -6 = 0, x = 6

solve x^2 +4 = 0
x^2 = -4
x = sqrt(-4) 
x = 2i, -21

The answer is B) 2i, 6, -6
andrezito [222]3 years ago
5 0
f(x)=x^4-32x^2-144\\\\f(x)=0\iff x^4-32x^2-144=0\\\\x^4-36x^2+4x^2-144=0\\\\x^2(x^2-36)+4(x^2-36)=0\\\\(x^2-36)(x^2+4)=0\iff x^2-36=0\ \vee\ x^2+4=0\\\\x^2=36\ \vee\ x^2=-4\\\\x=\pm\sqrt{36}\ \vee\ x=\pm\sqrt{-4}\\\\x=-6\ \vee\ x=6\ \vee\ x=-2i\ \vee\ x=2i

Answer: <span>C) 2i, 6, -6</span>

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Points:(2,3) and (-4,15) <br><br> I need help please
liberstina [14]

The equation intersecting these two points is:

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8 0
3 years ago
(Score for Question 1: ___ of 5 points)
Norma-Jean [14]

Answer:

The separated values of x and y from the given equation is x=\frac{13-3y}{4}, y=1-\frac{4}{13}x

Step-by-step explanation:

Given equation is 4 x+3y=13

Now to find the separated values of x and y with the given equation.

Now find the value of x and so we have to separate the value x from the given equation

4x+3y=13

4x=13=3y

x=\frac{13-3y}{4}

Therfore x=\frac{13-3y}{4}

Now find the value of y and so we have to separate the value y from the given equation

4x+13y=13

13y=13-4x

y=\frac{13-4x}{13}

y=\frac{13}{13}-\frac{4x}{13}

y=1-\frac{4}{13}

Therefore y=1\frac{4}{13}x

The separated values of x and y from the given equation is

x=\frac{13-3y}{4}, y=1-\frac{4}{13}x

7 0
4 years ago
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melisa1 [442]

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B

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