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Verizon [17]
2 years ago
9

PLEASE HELP!!! WILL MARK BRAINLIEST!!!!!

Mathematics
1 answer:
just olya [345]2 years ago
8 0

The piecewise defined function is

f(t) = 800 * 1.5^t for 0 - 2

f(t) = 2700 * 1.05^t-3 for 3 to 9

<h3>How to determine the piecewise functions?</h3>

On the domain of t = 0 to 2, we have:

f(0) = 800

f(1) = 1200

f(2) = 1800

The above function is a piecewise function with the following definition

f(0) = 800 --- initial value

r = 1200/800 = 1.5 --- rate

The function is then represented as:

f(t) = f(0) * r^t

This gives

f(t) = 800 * 1.5^t

On the domain of t = 3 to 9, we have:

f(3) = 2700

f(4) = 2835

The above function is a piecewise function with the following definition

f(3) = 800 --- initial value

r = 2835/2700 = 1.05 --- rate

The function is then represented as:

f(t) = f(3) * r^t-3

This gives

f(t) = 2700 * 1.05^t-3

Hence, the piecewise defined function is

f(t) = 800 * 1.5^t for 0 - 2

f(t) = 2700 * 1.05^t-3 for 3 to 9

Read more about piecewise defined function at

brainly.com/question/26145479

#SPJ1

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9 - 6 + 4 × 13

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Zina [86]

Answer:

\displaystyle \frac{1}{2}(x^3-2x^2-5x+6)

Step-by-step explanation:

<u>Polynomials </u>

a)

The polynomial whose graph is shown is of third degree because it has three real roots. The roots of a polynomial are the values of x that make the expression equal to zero. We can see it happens three times in the graph provided. The roots or zeros are  

x=-2, x=1, x=3

b)

The factored form of a polynomial whose roots x_1, x_2, x_3 are known is

a(x-x_1)(x-x_2)(x-x_3)

We know the value of the roots, thus the polynomial is written as

a(x+2)(x-1)(x-3)

We need to find the value of a. We do that by replacing the value of x=0 and finding a that makes f(0)=3 (as seen in the graph). Thus

a(0+2)(0-1)(0-3)=3

a(2)(-1)(-3)=3

a(6)=3

\displaystyle a=\frac{3}{6}

\displaystyle a=\frac{1}{2}

Thus the factored form of the polynomial is

\displaystyle \frac{1}{2}(x+2)(x-1)(x-3)

c)

Let's multiply all the factors

\displaystyle \frac{1}{2}(x+2)(x-1)(x-3)

\displaystyle \frac{1}{2}(x^2+2x-x-2)(x-3)

\displaystyle \frac{1}{2}(x^2+x-2)(x-3)

\displaystyle \frac{1}{2}(x^3-3x^2+x^2-3x-2x+6)

\boxed{ \displaystyle \frac{1}{2}(x^3-2x^2-5x+6)}

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Answer:

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Answer:

38

Step-by-step explanation:

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