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enyata [817]
3 years ago
15

Find the root(s) of f(x) = (x + 5)2(x - 9)2(x + 1).

Mathematics
1 answer:
Keith_Richards [23]3 years ago
7 0

Answer:

x = -5,-1, 9. -1 with multiplicity 1, 9 with multiplicity 1, and -5 with multiplicity 1.

Step-by-step explanation:

the coefficients do not change the roots of the equation but rather the local minima and maxima, and Minimum or Maximum depending on the function.

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Make a table showing all the possible outcomes for spinning the pointer of a spinner with four equal-sized sections labeled 1–4
natka813 [3]

Answer:

12.5%

Step-by-step explanation:

There are two types of spinners here and the outcome of them is independent. That means

P(odd number,C) = P(odd number) * P(C)

There are two odd numbers out of four numbers in the first spinner. The chance of odd number will be:

P(odd) = 2/(2+2)= 1/2= 50%

There are four letters and the desired outcome is C. The chance for C will be:

P(C)= 1/4= 25%

Then the chance will be:

P(odd number,C) = P(odd number) * P(C)

P(odd number,C) =50% * 25% = 12.5%

3 0
3 years ago
Help!!! Cubes! Hurry ima fail
aksik [14]

Answer:

your answer is 9

Step-by-step explanation:

Just trust me!

4 0
3 years ago
Read 2 more answers
Which pair shows equivalent expressions?
HACTEHA [7]

Answer:

18/6=324/6=418/24=3/4

Step-by-step explanation:

if you multiply 2 times something you can’t get anything 11. the closest denominator is 5 and 5x2=10 so you have one left to get 11 so the answer is =5with 1$ still left. maria can buy 5 fish but she will have 1$ left over.

Hope this helps.

8 0
3 years ago
for any positive integer n, the sum of the first n positive integers equals n(n 1)/2. what is the sum of all the even integers b
marusya05 [52]

The sum of all the even integers between 99 and 301 is 20200

To find the sum of even integers between 99 and 301, we will use the arithmetic progressions(AP). The even numbers can be considered as an AP with common difference 2.

In this case, the first even integer will be 100 and the last even integer will be 300.

nth term of the AP = first term + (n-1) x common difference

               ⇒    300 = 100 + (n-1) x 2

Therefore, n = (200 + 2 )/2 = 101

That is, there are 101 even integers between 99 and 301.

Sum of the 'n' terms in an AP = n/2 ( first term + last term)

                                                = 101/2 (300+100)

                                                = 20200

Thus sum of all the even integers between 99 and 301 = 20200

Learn more about arithmetic progressions at brainly.com/question/24592110

#SPJ4        

7 0
1 year ago
Please help me if possible
ss7ja [257]

Answer: 8 cm

Step-by-step explanation:

To do this, we just need to divide 256 by each of the numbers.

First let’s divide 256 by 8

256/8=32

Then divide 32 by 4

32/4 = 8.

Therefore your answer is 8 cm.

3 0
3 years ago
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