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4 years ago

Given f(x) = 3 and g(x) = 5x + 7, find (f + g)(-8).

Mathematics

1 answer:

levacccp [35]4 years ago

4 0

Answer:

-30

Step-by-step explanation:

f(x) = 3 so f(-8) = 3;

g(x) = 5x + 7, so g(-8) = 5(-8) + 7 = -33

Then (f + g)(-8) = 3 - 33 = -30

This is the sum of two functions both evaluated at x = -8.

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A bag contains 4 yellow marbles and 6 green marbles. Two marbles are drawn at random. One marble is drawn and not replaced. Then

Pie

$\frac{6}{10} \times \frac{4}{9} \\ = \frac{24}{90} \\ = \frac{4}{15}$

Answer is 2nd option, 4/15

Hope this helps. - M

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3 years ago

the vertex angle of an isosceles triangle has a measure of 24 degrees. what are the measures of the base angles?

svlad2 [7]

First you must remember that: An isosceles triangle consists of a vertex angle and 2 CONGRUENT base angles.

subtract 180-24
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4 years ago

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

Thepotemich [5.8K]

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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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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