The audio-visual technician for a hotel ordered 4 boxes of microphone cords. There are 12 cords in each box.

2 answers:

4 0

Your equation to solve is 108-(12×4).
This is 108-48 which is 60.
There were 60 cords at the hotel before the order. Since you need to find the boxes you do 60/12 which is 5.
There were 5 boxes.

Pavel [41]3 years ago

4 0

Answer: 5 boxes.

Step-by-step explanation:

Given : The audio-visual technician for a hotel ordered 4 boxes of microphone cords.

There are 12 cords in each box.

Then, the number of cords in 4 boxes will be $4\times12=48$

If there are now a total of 108 cords at the hotel, then the number of corfds were before the order : 108-48=60

Then, the number of boxes were in hotel before the order :-

$=\dfrac{60}{12}=5$

Hence, there were 5 boxes of cords in the hotel before the order.

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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.)$