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liubo4ka [24]
2 years ago
9

W/5+20=22. How do I do this

Mathematics
2 answers:
V125BC [204]2 years ago
7 0

Answer:

W = 10

Step-by-step explanation:

W / 5 + 20 = 22

W / 5 - 20 = 22 - 20

W / 5 = 2

W * 5 = 2 * 5

W = 10

Tanzania [10]2 years ago
3 0

Answer:

Make me as brainliest plz

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The weight of an organ adult male has a bell-shaped distribution with a mean of 330
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Which number is less than 0.37? A. 0.72 B. 0.5 C. 0.35 D. 0.41
ivolga24 [154]

Answer:

0.35 is less than 0.37 because 35 < 37.

8 0
3 years ago
Please answer! I'm so close to getting an A in my class, I believe this assignment can get me there, but I'm stuck on this probl
skelet666 [1.2K]
The Answer is d Because x is 2 and y is 1 in both "problems"

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Over a period of 12hours the temperature dropped 36° ° F. What was the average change in temperature in one hour?
mote1985 [20]

divide the temperature by the number of hours

36/12

answer: 3 degrees

8 0
3 years ago
Air is being lost from a spherical balloon at a constant rate of 3/2cm3s-1. Find the rate at which the radius is decreasing at t
cricket20 [7]

Answer:

The rate at which the radius is decreasing when the radius is 6 cm is approximately 3.316 × 10⁻³ cm/s

Step-by-step explanation:

The rate at which air is being lost from the balloon = 3/2 cm³/s

The rate at which the radius is decreasing when the radius is 6 cm long is given as follows;

The rate at which air is being lost from the balloon = dV/dt = 3/2 cm³/s

dV/dt = dV/dr × dr/dt

Where;

dr/dt = The rate at which the radius is decreasing

dV/dr = d(4/3×π×r³)/dr = 4·π·r²

Therefore, we have;

dr/dt = (dV//dt)/(dV/dr) = (3/2 cm³/s)/(4·π·r²)

dr/dt = (3/2 cm³/s)/(4·π·r²)

When r = 6 cm, we have;

dr/dt = (3/2 cm³/s)/(4 × π × (6 cm)²) ≈ 3.316 × 10⁻³ cm/s

Therefore, the rate at which the radius is decreasing, dr/dt, when the radius is 6 cm long ≈ 3.316 × 10⁻³ cm/s.

7 0
3 years ago
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