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grigory [225]
3 years ago
8

Algebra 2A Help ! Tadakimasu ~

Mathematics
1 answer:
Ludmilka [50]3 years ago
7 0
Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)

first factor into (x-r1)(x-r2)... form

p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)

multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2 

so

p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the  root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real

baseically

(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the  root 1 and has multiplity 1

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schepotkina [342]

Answer:

9 + 10 = 21

Step-by-step explanation:

9 + 10 = 21

Factor out 9 and 10

9 = 3 · 3   10 = 2 · 5

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4 0
3 years ago
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$1500 is invested at a rate of 3% compounded monthly. Write a compound interest function to model this situation. Then find the
Marianna [84]

Answer:

<u>Equation</u>:  F=1500(1.0025)^{12t}

<u>The balance after 5 years is:  $1742.43</u>

<u></u>

Step-by-step explanation:

This is a compound growth problem . THe formula is:

F=P(1+\frac{r}{n})^{nt}

Where

F is future amount

P is present amount

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n is the number of compounding per year

t is the time in years

Given:

P = 1500

r = 0.03

n = 12 (compounded monthly means 12 times a year)

The compound interest formula modelled by the variables is:

F=1500(1+\frac{0.03}{12})^{12t}\\F=1500(1.0025)^{12t}

Now, we want balance after 5 years, so t = 5, substituting, we get:

F=1500(1.0025)^{12t}\\F=1500(1.0025)^{12*5}\\F=1500(1.0025)^{60}\\F=1742.43

<u>The balance after 5 years is:  $1742.43</u>

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(5,26)

Step-by-step explanation:

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