Plz help, And no links please

1 answer:

6 0

I did not get any zeros since the graph doesn’t cross the x axis, meaning that there are no rational zeros

However, here is the method u can use to find the zeros lol

You can use the quadratic formula in order to get the zeros

This is the equation therefore use these values
ax^2+bx+c=0

A=1
B= -5
C=12

The quadratic formula is -b±√(b^2-4ac))/2a (I left a picture just in case)

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I need answer for this quick please

Anna007 [38]

Answer:

- 6

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

$\frac{f(b)-f(a)}{b-a}$

Here [ a, b ] = [ - 1, 4 ] , then

f(b) = f(4) = 4³ - 6(4)² - 4 = 64 - 96 - 4 = - 36

f(a) = f(- 1) = (- 1)³ - 6(- 1)² - (- 1) = - 1 - 6 + 1 = - 6 , then

average rate of change = $\frac{-36-(-6)}{4-(-1)}$ = $\frac{-36+6}{4+1}$ = $\frac{-30}{5}$ = - 6

5 0

2 years ago

Describe the steps you would use to factor 2x3 + 5x2 – 8x – 20 completely. Then state the factored form.

WARRIOR [948]

The polynomial $2x^3 + 5x^2-8x-20$ may have solutions which are the divisors of -20, therefore -20 has the following divisors: $\pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20$ .
If x=1, then $2\cdot1^3 + 5\cdot1^2-8\cdot1-20=-20\neq 0$ ,
if x=-1, then $2\cdot(-1)^3 + 5\cdot(-1)^2-8\cdot(-1)-20=-9\neq 0$ ,
if x=2, then $2\cdot2^3 + 5\cdot2^2-8\cdot2-20=0$ , then x=2 is a solution and you have the first factor (x-2).
If x=-2, then $2\cdot(-2)^3 + 5\cdot(-2)^2-8\cdot(-2)-20=0$ , then x=-2 is a solution, so you have the second factor (x+2).
Since x-2 and x+2 are two factors of $2x^3 + 5x^2-8x-20$ , then the polynomial $x^2-4$ is a divisor of $2x^3 + 5x^2-8x-20$ and dividing the polynomial $2x^3 + 5x^2-8x-20$ by $x^2-4$ you obtain
$2x^3 + 5x^2-8x-20=(x-2)(x+2)(2x+5)$ .

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2 years ago

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What are the missing numbers<br> ?, 3.2, 2.4, ?, 0.8

UNO [17]

The first missing number is 4
The second missing number is 1.6

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2 years ago

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Ronch [10]

B. 1 angle A congruent angle B
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