(Sorry for the spam, making sure these are right) One zero of the polynomial function f(x)=x^3-9x^2+20x is x=0. What are the polynomial function? Choices:

Answers

Answer 1

Answer:

Replace f(x) with 0 and solve for x. We do this because the x intercepts always occur when y = 0. Keep in mind that y = f(x).

f(x)=x^3-9x^2+20x

0=x^3-9x^2+20x

x^3-9x^2+20x = 0

x(x^2-9x+20) = 0 .... factor out GCF x

x(x-5)(x-4) = 0 ... factor the stuff inside

x = 0 or x-5 = 0 or x-4 = 0 ... zero product property

x = 0 or x = 5 or x = 4

The roots or zeros are 0, 5, 4

Answer: Choice D

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Find the slope between the two points (-5,5) and (3,-5)

Answers

the formula is y2 minus y1 over x2 minus x1 or rise (y) over run (x). So you would do (-5) - (5) over (3) - (-5). that would equal -10 over 8, or -5/4 when you simplify. your slope would be -5/4

What is 5-1???????????????????????????????

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Answer : 4

I hope that's help !

Oh i know this, its 4!!

Please help asap 25 pts

Answers

Your answer is the first one which is the graph is of exponential growth

It is an exponential growth!

9(n-4)-7n=32-2(n+8)

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Answer:

answer is 13

Step-by-step explanation:

hope it helps.

What is the radius of a circle with center (2,1) and containing point (0,1)

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Answer:

The radius is 2 or The simplified radius is 4 ( if needed).

Step-by-step explanation:

- The equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$ where the center is $(h,k)$ , the point as $(x,y)$ and the radius $r^2$ .

-Put both the center and the point onto that equation:

$(0-2)^2+(1-1)^2=r^2$

-Then, solve them up:

$(0-2)^2+(1-1)^2 =r^2$

$(-2)^2+0=r^2$

$4 + 0=r^2$

$4 = r^2$

$√(4) =√(r^2)$

$2=r$

Look at this sequence.95, 80, 65, 50,... Which function represents this sequence? A. a(n) = 110n - 15 B. a(n) = 15n + 80 C. a(n) = 95n - 15 D. a(n) = -15n + 95

Answers

Answer:

Hi,
answer A

Step-by-step explanation:

The sequence is an AP with first term equal 95 and common difference equal -15

$a_n=a_1+(-15)*(n-1)= 95+(-15)*(n-1)=95-15n+15\n\n\boxed{a_n=110-15n}\n$

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