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astra-53 [7]
3 years ago
8

Find the area of a regular pentagon with the apothem of 24.3 cm and a side length of 35.3

Mathematics
1 answer:
steposvetlana [31]3 years ago
8 0

Answer: A=2144\ cm^2

Step-by-step explanation:

For this exercise you need  to use the following formula for calculate the area of regular polygon:

A=\frac{s^2n}{4tan(\frac{180}{n})}

Where is "s" the length of any side, "n" is the number of sides.

Int this case you know that it is regular pentagon, which means that it has five sides. Then you can identify that the values of "n" and "s":

s=35.3cm\\n=5

 Therefore, substituitng values into the formula, you get that the area f the pentagon is:

A=\frac{(35.3cm)^2(5)}{4tan(\frac{180}{5})}\\\\A=2144\ cm^2

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4/3 per minute.

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Multiply both 1/4 and 1/3 to get 1 minute. 1/3x4=4/3

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3 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

so

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
Express 0.96 as a fraction
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Since 6 is in the hundredths place,

96/100

You can divide that by two for a different, but equal fraction.

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