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2 years ago

Find the product of (4x − 3)(2x2 − 7x 1). 8x3 − 22x2 17x− 3 8x3 8x2 4x − 3 8x3 − 34x2 25x − 3 8x3 − 42x2 25x − 3.

1 answer:

3 0

Product of two polynomials can be done by distribution of multiplication over addition. The product of $(4x -3)(2x^2 -7x + 1)$ is $8x^3 - 34x^2 + 25x - 3$

<h3>What is distributive property of multiplication over addition?</h3>

Suppose a, b and c are three numbers. Then we have:

$a(b + c) = a\times b + a\times c$

(a(b+c) means a multiplied to (b+c). The sign of multiplication is usually hidden when using symbols and both quantities which are in multiplication are written together without space)

The product result of the given expression is obtained as:

$(4x -3)(2x^2 -7x + 1) = 4x(2x^2 - 7x + 1) -3(2x^2 - 7x + 1)\\(4x -3)(2x^2 -7x + 1) = 8x^{1 + 2} - 28x^2 + 4x - 6x^2 + 21x - 3\\(4x -3)(2x^2 -7x + 1) = 8x^3 + (-28-6)x^2 + (4 + 21)x - 3\\(4x -3)(2x^2 -7x + 1) = 8x^3 - 34x^2 + 25x - 3$

Thus,

The product of $(4x -3)(2x^2 -7x + 1)$ is $8x^3 - 34x^2 + 25x - 3$

Learn more about polynomial product here;

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The complete question in the attached figure

Let
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The graph h = −16t^2 + 25t + 5 models the height and time of a ball that was thrown off of a building where h is the height in f

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Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

$h(t)=-16t^{2}+25t+5$

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

Factor -16

$h(t)=-16(t^{2}-\frac{25}{16}t)+5$

Complete the square

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}$

$h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}$

Rewrite as perfect squares

$h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

The vertex is the point $(\frac{25}{32},\frac{945}{64})$

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

$0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}$

$16(t-\frac{25}{32})^{2}=\frac{945}{64}$

$(t-\frac{25}{32})^{2}=\frac{945}{1,024}$

square root both sides

$(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}$

$t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}$

the positive value is

$t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec$

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