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3 years ago

Evaluate the expression for the given value of the variable. |4b-8|+|-1-b^2|+2b^3;b=2

Mathematics

1 answer:

Elena-2011 [213]3 years ago

4 0

In this question we already know the value of 'b' so put the value b=2 in the equation

=|4(2)-8| + |-1-(2)^2| + 2(2)^3

=|8-8| + |-1-4| + 2(8)

=0 -5 +16

=11

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Hi there! :)

Answer:

In order to maintain the same ratio, she should sell 18 band posters.

Step-by-step explanation:

Initial ratio:

Ratio is: 105 movie posters for 90 band posters = 105 : 90

She sells 21 movie posters, which means that we are left with 105 movies minus the 21 she sold.

105 - 21 = 84 → She now has 84 movie posters.

In order to answer your question, you'll need to use the cross product method:

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3 years ago

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$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

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(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

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We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

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We plug in 0 for x and 1 for x

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$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

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