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2 years ago

What is the domain of the function graphed below?

Mathematics

2 answers:

grigory [225]2 years ago

5 0

It is 7,infinite charater thing

Vlad1618 [11]2 years ago

3 0

Answer:

The domain is the set of x values for which the function resides in.

In this graph, it is unclear as to if the function goes on to infinity, but if it does, the answer is quite easily:

(-2, 4] and [7, ∞)

since the function starts on the left at -2 then continues to the right, with a pause, then indefinitely after.

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I need help please. Thank you.

AfilCa [17]

The answers are C and B.
Answer A has infinite solutions, and D has no solutions

8 0

2 years ago

Approximately 0.8 of middle school students have cell phones. Which of the following fractions describes how many middle school

lana [24]

Answer: E. 4/5

Step-by-step explanation: You have to put the decimal (0.8) over its place value. So in this case 0.8 turns into 8/10. Then you must simplify 8/10, by dividing the numerator (8) and the denominator (10) by the greatest common factor (2). Which gives you the answer 4/5.

8 0

2 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

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

8 0

2 years ago

Is x+2x^3/x^2 a polynomial true or false

Elena L [17]

Answer:True,because there are two or more variable/numbers in there

Step-by-step explanation:

6 0

2 years ago

20 points! Will mark the brainliest!

andrew-mc [135]

$\dfrac{ \dfrac{x+3}{4x^2-16} }{ \dfrac{2x^2+10x+12}{2x-4} }$

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Write the divide fraction horizontally:
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$= \dfrac{x+3}{4x^2-16} \div \dfrac{2x^2+10x+12}{2x-4}$

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Factorise the numerators and denominators when possible:
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$= \dfrac{x+3}{4(x+2)(x-2)} \div \dfrac{ 2(x + 3) (x + 2)}{2(x-2)}$

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Convert the divide fraction to multiplication fraction
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$= \dfrac{x+3}{4(x+2)(x-2)} \times \dfrac{2(x-2)}{2(x+3)(x+2)}$

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Cancel the factors
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$= \dfrac{1}{4(x+2)} \times \dfrac{1}{(x+2)}$

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Combine to single fraction
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$= \dfrac{1}{4(x+2)^2}$

4 0

3 years ago