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eduard
2 years ago
9

Does anybody have a REAL answer for this?

Mathematics
1 answer:
S_A_V [24]2 years ago
6 0
Area of square = 15 x 15 = 225 cm^2
area of circle = pi r^2 = (3.14)(15)(15) = 706.5 cm^2

1/4 of circle = 706.5 / 4 = 176.6

so area of yellow region = 225  - 176.6 = 48.4 cm^2

answer
48.4 cm^2
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Please help ASAP help please due soon
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Answer:

75 in^2

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The surface area is the sum of the areas

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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)

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so

The x-coordinate of the vertex represent the time when the ball reach the maximum

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Convert the equation in vertex form

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h(t)=-16(t^{2}-\frac{25}{16}t)+5

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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})

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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

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the positive value is

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

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