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skelet666 [1.2K]

3 years ago

5

A seagull is gliding 53 feet above the surfaces of the water. It then dives 70 feet to catch a fish. How many feet below the sur

face of the water does the seagull dive?

1 answer:

Morgarella [4.7K]3 years ago

6 0

Answer:

17 feet

Step-by-step explanation:

the seagull is originally 53 feet above water and dives 70 feet down. so 53-70=-17. 17 feet below water.

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Find the final value and the compound interest if $8000 is invested for 2 years at 27% p.a.

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Find AB. Round to the nearest tenth if necessary.<br>

shusha [124]

Answer:

I think the right answer would be 7

Step-by-step explanation:

We got-ta use the equation: CA * BA = DA²

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3 0

3 years ago

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min, and its coarseness is such that it forms a pile in the shap

11Alexandr11 [23.1K]

Answer:

0.25 feet per minute

Step-by-step explanation:

Gravel is being dumped from a conveyor belt at a rate of 20 ft3/min. Since we are told that the shape formed is a cone, the rate of change of the volume of the cone.

$\dfrac{dV}{dt}=20$ ft^3/min$

$\text{Volume of a cone}=\dfrac{1}{3}\pi r^2 h$

Since base diameter = Height of the Cone

Radius of the Cone = h/2

Therefore,

$\text{Volume of the cone}=\dfrac{\pi h}{3} (\dfrac{h}{2}) ^2 \\V=\dfrac{\pi h^3}{12}$

$\text{Rate of Change of the Volume}, \dfrac{dV}{dt}=\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}$

Therefore: $\dfrac{3\pi h^2}{12}\dfrac{dh}{dt}=20$

We want to determine how fast is the height of the pile is increasing when the pile is 10 feet high.

$h=10$ feet$\\\\\dfrac{3\pi *10^2}{12}\dfrac{dh}{dt}=20\\25\pi \dfrac{dh}{dt}=20\\ \dfrac{dh}{dt}= \dfrac{20}{25\pi}\\ \dfrac{dh}{dt}=0.25$ feet per minute (to two decimal places.)$

When the pile is 10 feet high, the height of the pile is increasing at a rate of 0.25 feet per minute

5 0

3 years ago