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natta225 [31]
3 years ago
7

Answer this pls, only need an answer A, B, C, or D

Mathematics
1 answer:
Butoxors [25]3 years ago
7 0

Answer:

it is the NOL and KLJ i didnt know how you were ordering the A B C or D

Step-by-step explanation:

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Solve the division equation using the model. (2 points)
julia-pushkina [17]
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Explanation
Factor the numerator and denominator and cancel the common factors.
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3 years ago
The polynomial of degree 4, P ( x ) , has a root of multiplicity 2 at x = 1 and roots of multiplicity 1 at x = 0 and x = − 2 . I
ICE Princess25 [194]

We want to find a polynomial given that we know its roots and a point on the graph.

We will find the polynomial:

p(x) = (183/280)*(x - 1)*(x - 1)*(x + 2)*x

We know that for a polynomial with roots {x₁, x₂, ..., xₙ} and a leading coefficient a, we can write the polynomial equation as:

p(x) = a*(x - x₁)*(x - x₂)...*(x - xₙ)

Here we know that the roots are:

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  • x = 0
  • x = -2

Then the roots are: {1, 1, 0, -2}

We can write the polynomial as:

p(x) = a*(x - 1)*(x - 1)(x - 0)*(x - (-2))

p(x) = a*(x - 1)*(x - 1)*(x + 2)*x

We also know that this polynomial goes through the point (5, 336).

This means that:

p(5) = 336

Then we can solve:

336 = a*(5 - 1)*(5 - 1)*(5 + 2)*5

336 = a*(4)*(4)*(7)*5

336 = a*560

366/560 = a = 183/280

Then the polynomial is:

p(x) = (183/280)*(x - 1)*(x - 1)*(x + 2)*x

If you want to learn more, you can read:

brainly.com/question/11536910

5 0
3 years ago
PLEASE HELP WILL GIVE BRANILYEST
Ivahew [28]

No because to get a proportional relationship you have to add the same number to all x to get y

8 0
3 years ago
The population of bacteria in a Petri Dish is growing at a rate of 0.8t^3 + 3.5 thousand per hour. Find the total increase in ba
aliina [53]

Answer:

p=9900\\ bacterias in the initial two hours

Step-by-step explanation:

the growing rate is given by the ecuation

p(t)=0.8(t)^{3} +3.5 thousand per hour

for t=2 we have

p(2)=0.8(2)^{3} +3.5 = 9.9 thousand

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In two hours we have 9900 bacterias

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3 years ago
Solve 2cos x+2cos 2x=0 on the interval [0,2pi)
rjkz [21]
2cosx+2cos2x=0 \\ -2sin^2x+2cos^2x+2cosx=0 \\ -2(1-cos^2x)+2cos^2x+2cosx=0 \\ -2+2cos^2x+2cos^2x+2cosx=0 \\ 4cos^2x+2cosx-2=0 \\ t=cos x \\ 4t^2+2t-2=0 \\ \Delta=4+32=36 \\ t_1= \frac{-2-6}{8}=-1 \\ t_2= \frac{-2+6}{8}= \frac{1}{2}  \\ cos x=-1 \lor cos x= \frac{1}{2}    \\ x=-\pi \lor x= \frac{5\pi}{3}
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3 years ago
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