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weeeeeb [17]
3 years ago
13

What is the approximate area of the circle shown below 9.4 cm

Mathematics
1 answer:
aivan3 [116]3 years ago
6 0

Answer:

Step-by-step explanation:

A= πr²

A= 22/7 multiply by the radius.

Divide 9.4cm by 2.

Whatever you get is your answer. Now take the formula I have shown you above. Whatever you get is your answer.

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Explain how to compare two ratios?
djverab [1.8K]

To compare ratios, write them as fractions. The ratios are equal if they are equal when written as fractions.
3 0
3 years ago
PROBLEM 2 The height of a golf ball in feet can be determined using the formula h(t) = -16t2 + 100x , where t is the number of s
wolverine [178]

Answer:

<h3>H. 150 ft</h3>

Step-by-step explanation:

Given the height of a golf ball in feet modelled by the formula h(t) = -16t² + 100x where t is in seconds, we are to find the height of the golf after 2.5 seconds. To do that we will simply substitute t = 2.5secs into the formula given;

h(t) = -16t^2 + 100x\\\\h(2.5) = -16(2.5)^2 + 100(2.5)\\\\h(2.5) = -16(6.25) + 250\\\\h(2.5) = -100+250\\\\h(2.5) = 150\\\\

Hence the height of the golf after 2.5 seconds is 150ft.

8 0
3 years ago
Which of the following equations are true for the number 9? Select all that apply.
telo118 [61]

Answer:

B and C

Step-by-step explanation:

1/9 = 1÷9

7÷9 = 7/9

Option B and C is the only equation true for the number 9

3 0
3 years ago
"We might think that a ball that is dropped from a height of 15 feet and rebounds to a height 7/8 of its previous height at each
tatyana61 [14]

Answer:

Total Time = 4.51 s

Step-by-step explanation:

Solution:

- It firstly asks you to prove that that statement is true. To prove it, we will need a little bit of kinematics:

                             y = v_o*t + 0.5*a*t^2

Where,   v_o : Initial velocity = 0 ... dropped

              a: Acceleration due to gravity = 32 ft / s^2

              y = h ( Initial height )

                             h = 0 + 0.5*32*t^2

                             t^2 = 2*h / 32

                             t = 0.25*√h   ...... Proven

- We know that ball rebounds back to 7/8 of its previous height h. So we will calculate times for each bounce:

1st : 0.25*\sqrt{15}\\\\2nd: 0.25*\sqrt{15} + 0.25*\sqrt{15*\frac{7}{8} } + 0.25*\sqrt{15*\frac{7}{8} } = 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} }\\\\3rd: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 2*0.25*\sqrt{15*(\frac{7}{8} })^2\\\\= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2\\\\4th: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 2*0.25*\sqrt{15*(\frac{7}{8} })^3 \\\\

= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 0.5*\sqrt{15*(\frac{7}{8} })^3

- How long it has been bouncing at nth bounce, we will look at the pattern between 1st, 2nd and 3rd and 4th bounce times calculated above. We see it follows a geometric series with formula:

  Total Time ( nth bounce ) = Sum to nth ( \frac{1}{2}*\sqrt{15*(\frac{7}{8})^(^i^-^1^) }  - \frac{1}{4}*\sqrt{15})

- The formula for sum to infinity for geometric progression is:

                                   S∞ = a / 1 - r

Where, a = 15 , r = ( 7 / 8 )

                                   S∞ = 15 / 1 - (7/8) = 15 / (1/8)

                                   S∞ = 120

- Then we have:

                                  Total Time = 0.5*√S∞ - 0.25*√15

                                  Total Time = 0.5*√120 - 0.25*√15

                                  Total Time = 4.51 s

5 0
3 years ago
PLEASE HELPPP!!!<br> .<br> .<br> I’m not sure how to do thiss
Marrrta [24]

let x be .24 repeating

100x = 24.24 repeating

x = .24 repeating

99x = 24

x = 24/99

Repeat this with the rest of the problems

7 0
3 years ago
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