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3 years ago

What algebraic expression is equivalent to this expression? 9(2x+6) - 18

Mathematics

2 answers:

pogonyaev3 years ago

8 0

Answer:

18x+36

Step-by-step explanation:

We are given the expression:

$9(2x+6)-18$

We want to simplify the expression. First, let's distribute the 9. Multiply each term inside the parentheses by 9.

$(9*2x)+(9*6)-18$

$(18x)+(54)-18$

Combine like terms. There are 2 constants (terms without variables): 54 and -18. We can add them together.

$(18x)+(54+-18)$

$(18x)+(54-18)$

$18x+36$

The expression 9(2x+6)-18 is equivalent to 18x+36

Svet_ta [14]3 years ago

3 0

Answer:

18x+36

Step-by-step explanation:

9(2x+6) - 18

Distribute

9*2x + 9 *6 -18

18x +54 -18

Combine like terms

18x+36

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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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