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Zinaida [17]
3 years ago
5

F(x)= -3x^3-x^2+x-3 identify the end behavior of each polynomial function

Mathematics
1 answer:
sashaice [31]3 years ago
3 0

end behavior is affect by the exponent and whether its negative or positive

y = y^3 starts negative from -∞ to 0 the from 0 to +∞ goes positive

since this equation is negative its switched

-∞ to 0  its positive x>0

0 to +∞ is negative x<0

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4. Stella says the equation x^2−8x+y^2+2y=5 has a center of (4,−1) and a radius of 5. Is she correct? Why or why not?
Ostrovityanka [42]
<h2>Stella is correct about center and wrong about radius      </h2>

Step-by-step explanation:

Equation of circle with (h,k) as center and radius r is given by,

          (x-h)²+(y-k)² = r²

Here equation is

          x²−8x+y²+2y=5

Changing in to (x-h)²+(y-k)² = r² form

           x²−8x + 16 - 16 +y²+2y+1 - 1=5  

           (x-4)² - 16 + (y+1)² -1 = 5

           (x-4)²  + (y+1)²  = 22

           (x-4)²  + (y+1)²  = 4.69²

           Center is (4,-1) and radius is 4.69

Stella is correct about center and wrong about radius            

8 0
2 years ago
Whats 12*5+55= <br><br> ill mark asa brainliest
Aleksandr-060686 [28]
First do PEMDAS
There's no parenthesis. So move on to exponent. There's also no exponents. Move on to  multiplication. There it is. 

12×5=60.

There's no division so now next one is addition. Hey look! There is addition. 

60+55=115.

So 115 is your answer.
5 0
3 years ago
Read 2 more answers
An instructor who taught two sections of engineering probability last term, the first with 20 students and thesecond with 30, de
Oliga [24]

Answer:

0.207

Step-by-step explanation:

This is an hypergeometric distribution problem

An hypergeometric distribution has the same sense as the discrete probabilities of binomial distribution, but unlike binomial distribution, hypergeometric distribution does not allow replacement.

Binomial distribution expresses the probability of picking k objects from n with replacement, but hypergeometric distribution expresses picking k objects from n without replacement, with the finite total population, N, containing K objects.

It is expressed mathematically as

h(k: n, K, N) = (ᴷCₖ)(ᴺ⁻ᴷCₙ₋ₖ)/(ᴺCₙ)

where

k = number of students in the 2nd section required to be in the first 15 graded projects (number of successes) = 10

n = total number of first graded projects (number of trials) = 15

K = number of students in the 2nd section of the class = 30

N = total number of students = 50

h(10: 15, 30, 50) = (³⁰C₁₀)(⁵⁰⁻³⁰C₁₅₋₁₀)/(⁵⁰C₁₅)

h(10: 15, 30, 50) = (³⁰C₁₀)(²⁰C₅)/(⁵⁰C₁₅)

= (30,045,015)(15,504)/(2,250,829,575,120)

P(X = 10) = 0.207

Hope this Helps!!!

7 0
3 years ago
Determine the values of X, Y and Z on the following number line:<br> HH<br> X -11 Y 9 Z
Dmitrij [34]
-11 y 17 with z and x on the line
3 0
3 years ago
HELP ASAP PLEASE!!!!!!
devlian [24]
A = (2+9)/2

a = 11/2

a = 5.5

__________________

b = (4+4)/2

b = 4

__________________

d = √(9-2)²+(4-4)²

d = √7²+0²

r = 3,5


____________________

(x-a)² + (y-b)² = r²

(x-5.5)² + (y-4)² = 3,5²

(x-5.5)² + (y-4)² = 12.25
5 0
3 years ago
Read 2 more answers
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