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3 years ago

What is the difference between 9 - 4 1/5

Mathematics

1 answer:

erica [24]3 years ago

3 0

Hey there!☺

$Answer:\boxed{4\frac{4}{5}}$

$Explanation:$

$9-4\frac{1}{5}$

Subtract the whole number by the fraction.

$\frac{24}{5}$

24/5 needs to be simplified so we will rewrite it like this:

$4\frac{4}{5}$

Hope this helps!

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A doormat is 2 feet long. If you placed 2 doormats in a line, would they be longer than 1 yard? Explain your answer.

tangare [24]

Answer:

Yes

Step-by-step explanation:

A yard is equivalent to 3 feet. If a singular doormat is 2 feet long then two doormats placed in a line would be 4 feet long. 4 > 3 meaning they would be longer than a yard.

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3 years ago

Elizabeth wants to write a letter to the editor of the school newspaper to

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Answer:

C. Almond Middle School, another local school, raised over $2,000

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Step-by-step explanation:

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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

3 years ago

PLEASEE HELP!! NUMBER 4!!

Illusion [34]

?????????????????????/ us a calculator to add up all the nub.

8 0

3 years ago

write a slope intercept equation for a line passing through the point (2,-2) that is parallel and perpendicular to the line x=-1

vekshin1

You would need two different lines to complete this as lines cannot be both parallel and perpendicular (these are opposites). The answers would be:

Parallel: x = 2

Perpendicular: y = -2

In order to find these, we first need to see that the original line of x = -1 is a horizontal line. Therefore, any line that is parallel should be horizontal as well. To get a horizontal line through the point (2, -2), the only option is x = 2.

Similarly, with the perpendicular line, if the original line is horizontal, the new line must be vertical. The only vertical line that goes through (2, -2) is y = -2.

5 0

3 years ago