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2 years ago

Write 1.888... as a mixed number

Mathematics

1 answer:

Leokris [45]2 years ago

3 0

Answer:

Step-by-step explanation:

The answer is 1-4/5

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PLEASE HELP WILL MARK BRAINLIEST!!!

Gala2k [10]

The Unit Rate is the start of the slope/the y-intercept of the slope.To find that we need to make the slope first.

(shown in the first picture)

We can see that the slope is a linear function/a straight line, and if we just follow the line down the slope we can see that it has a y-intercept/unit rate of 2/3 of 1 whole.

(the second picture)

6 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

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

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2 years ago

Evaluate: 54-75+81-(-27)+53

romanna [79]

Hi the answer is 344

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3 years ago

Read 2 more answers

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Rama09 [41]

Domain: {-5, 7}
Range: {-9, 4, 8}

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2 years ago

What is the circumference of this circle?

Yanka [14]

C ≈ 307.88
Hope that helps.

7 0

3 years ago