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3 years ago

Is x − 8 a factor of the function f(x) = −2x3 + 17x2 − 64? Explain.

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

5 0

Answer:

Yes

Step-by-step explanation:

I used synthetic division to find the answer. I added a photo of my work.

Download pdf

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-6x-(-8)=2 step by step

levacccp [35]

Equation/Question:

-6x-(-8)=2

Answer/Way I did it:

1. Remove parentheses.

-6x+8=2

2. Subtract 8 from both sides.

-6x=2-8

3. Simplify 2-8 to -6.

-6x=-6

4. Divide both sides by -6.

x=1

-------------

Checking:

Equation/Question:

−6x−(−8)=2

----------------------

1. Let x=1.

−6×1+8=2

2. Simplify -6×1 to −6.

-6+8=2

3. Simplify -6+8 to 2.

2=2

Done by NeighborhoodDealer

6 0

3 years ago

Write 1-3log5x as a single logarithm

anyanavicka [17]

1 can be written as log 10

so,
log 10 - 3 log 5x = log 10 - log (5x)^3
= log 10/(5x^3)

4 0

3 years ago

Read 2 more answers

PLS help ASAp only available until 11:59

sergij07 [2.7K]

Answer:

this is the answee

3 0

3 years ago

Find the unit rate

kaheart [24]

Answer:

a) 60 mph

b) 32 miles per gallon

c) $2.40

d) $0.80

e) 54 wpm

7 0

3 years ago

PLEASE HELP!!

Bezzdna [24]

Answer:

$y=-\dfrac{9}{40}x^2+\dfrac{441}{40}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}$

Step-by-step explanation:

If a parabola has its vertex on the y-axis, then its equation is

$y=ax^2+b$

This parabola passes through the point R(3,9), then

$9=a\cdot 3^2+b\\ \\9=9a+b$

The area of the right triangle PQR is

$A_{PQR}=\dfrac{1}{2}\cdot PQ\cdot QR$

Find PQ and QR, if $P(x_1,0),\ Q(3,0),\ R(3,9):$

$PQ=\sqrt{(x_1-3)^2+(0-0)^2}=|x_1-3|\\\\QR=\sqrt{(3-3)^2+(0-9)^2}=9$

Now,

$40.5=\dfrac{1}{2}\cdot |x_1-3|\cdot 9\\ \\90=9|x_1-3|\\ \\|x_1-3|=10\\ \\x_1-3=10\ \text{or}\ x_1-3=-10\\ \\x_1=13\ \text{or }x_1=-7$

We get two possible points $P_1(-7,0)$ and $P_2(13,0).$

For point $P_1:\\$

$0=a\cdot (-7)^2+b\\ \\49a+b=0$

So,

$b=-49a\\ \\9=9a-49a\\ \\-40a=9\\ \\a=-\dfrac{9}{40}\\ \\b=\dfrac{441}{40}\\ \\y=-\dfrac{9}{40}x^2+\dfrac{441}{40}$

For point $P_2:\\$

$0=a\cdot (13)^2+b\\ \\169a+b=0$

So,

$b=-169a\\ \\9=9a-169a\\ \\-160a=9\\ \\a=-\dfrac{9}{160}\\ \\b=\dfrac{1,521}{160}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}$

5 0

3 years ago