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inessss [21]
3 years ago
7

I need help plss!! I don’t understand

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0

Since it says what the area of the circle is, you need to work the problem backwards from finding the area. So it would be 100 ÷ 3.14 = 31.84 that is the radius. Next you need to find the diameter, so 31.84 × 2 = 63.68. Now you multiply 63.68 × 3.14 = 199.95, round that to 200, divide 200 ÷ 2 = 100 ÷ 2 = 50 miles

Answer= 50 miles

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Answer for brainilest and 30 points.
Karo-lina-s [1.5K]

Answer:

Top Right

Step-by-step explanation:

231/pi(3)^2 =8.17

Round down to 8

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2 years ago
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A swimming pool is 50 metres long and 25 metres wide. How many lengths of the pool will a competitor in a 1500 metre race have t
GarryVolchara [31]
<h3>Answer:  30</h3>

Reason:

I'm assuming that the swimmer travels along the 50 meter side

Divide 1500 over 50 to get 1500/50 = 30, which tells us that the swimmer must make 30 trips.

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rebecca's cell phone bill is $52 she pays a monthly fee of $20 and $0.05 per minute that she uses .How many minutes does rebecca
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She can only talk for 400 minutes or 6 hours and 40 minutes
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3 years ago
Object is thrown upward from a height of 15 ft at an initial vertical velocity of 30 ft per second. How long will it take to hit
dem82 [27]

Answer:

2.25 s.

Step-by-step explanation:

We'll begin by calculating the time taken for the object to get to the maximum height from the point of projection. This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₁) to reach the maximum height from the point of projection =?

v = u – gt₁ (since the object is going against gravity)

0 = 30 – (32.15 × t₁

0 = 30 – 32.15t₁

Collect like terms

0 – 30 = – 32.15t₁

– 30 = – 32.15t₁

Divide both side by – 32.15

t₁ = –30 / –32.15

t₁ = 0.93 s

Next, we shall determine the maximum height reached by the object from the point of projection.

This can be obtained as follow:

Initial velocity (u) = 30 ft/s

Final velocity (v) = 0 ft/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Maximum height (h) reached from the point of projection =?

v² = u² – 2gh (since the object is going against gravity)

0² = 30² – (2 × 32.15 × h)

0 = 900 – 64.3h

Collect like terms

0 – 900 = – 64.3h

– 900 = – 64.3h

Divide both side by – 64.3

h = –900 / –64.3

h = 14 ft

Thus, the maximum reached by the object from the point of projection is 14 ft.

Next, we shall determine the height to which the of object is located from the maximum height reached to the ground. This can be obtained as follow:

Height (h₀) from which the object was projected = 14 ft

Maximum Height (h) reached from the point of projection = 14 ft

Height (hₗ) to which the of object is located from the maximum to the ground =?

hₗ = h₀ + h

hₗ = 14 + 14

hₗ = 28 ft

Thus, the height to which the of object is located from the maximum reached to the ground is 28 ft.

Next, we shall determine the time taken for the object to get to the ground from the maximum height reached. This can be obtained as follow:

Height (hₗ) to which the of object is located from the maximum to the ground = 28 ft

Acceleration due to gravity (g) = 9.8 m/s² = 3.28084 × 9.8 = 32.15 ft/s²

Time (t₂) taken for the object to get to the ground from the maximum height reached =?

hₗ = ½gt₂²

28 = ½ × 32.15 × t₂²

28 = 16.075 × t₂²

Divide both side by 16.075

t₂² = 28 / 16.075

Take the square root of both side

t₂ = √(28 / 16.075)

t₂ = 1.32 s

Finally, we shall determine the time take for the object to get to the ground from the point of projection. This can be obtained as follow:

Time (t₁) to reach the maximum height from the point of projection = 0.93 s

Time (t₂) taken for the object to get to the ground from the maximum height reached = 1.32 s

Time (T) take for the object to get to the ground from the point of projection =?

T = t₁ + t₂

T = 0.93 + 1.32

T = 2.25 s.

Therefore, the time take for the object to get to the ground from the point of projection is 2.25 s.

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A because when you distribute you get 3.5t + 21 = 7. Then subtract 21 from both sides to get 3.5t = -14. Then divide both sides by 3.5t to get -4. 
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3 years ago
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