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2 years ago

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The cafeteria ordered 400 paper plates they use 226 at breakfast they bought 100 more then they use some plates for lunch now th

ere are 78 plates how many plates did they use at lunch

2 answers:

stiv31 [10]2 years ago

4 0

196 plates were used at lunch

xenn [34]2 years ago

3 0

196 plates were used for lunch.

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The volume of a spherical balloon is increasing at the constant rate of 8 cubic feet per minute. How fast is the radius incresin

OverLord2011 [107]

Answer:

Rate of increase of radius = 0.0064 ft/sec

Rate of increase of surface area = 1.61 $ft^2/sec$

Step-by-step explanation:

Given that:

Rate of change of volume of a spherical balloon = 8 cubic feet per minute

$\dfrac{dV}{dt} = 8 ft^3/min$

Radius, $r$ = 10 feet

To find:

The rate of change of radius at this moment and rate of change of surface area at this moment?

Solution:

First of all, let us have a look at the formula:

$1.\ V = \dfrac{4}{3}\pi r^3\\2.\ A =4\pi r^2$

Now, differentiating the volume and area, we get:

$\dfrac{dV}{dt} = \dfrac{4}{3}\times 3 \pi r^2 \dfrac{dr}{dt}\\\Rightarrow \dfrac{dV}{dt} = 4 \pi r^2 \dfrac{dr}{dt}$

$\dfrac{dA}{dt} = 4\pi \times 2 r \dfrac{dr}{dt}\\\Rightarrow \dfrac{dA}{dt} = 8\pi r \dfrac{dr}{dt}$

$\dfrac{dV}{dt} = 8 = 4\pi r^2 \dfrac{dr}{dt}\\\Rightarrow 8 = 4\times 3.14 \times 10^2 \dfrac{dr}{dt}\\\Rightarrow 2 = 3.14 \times 10^2 \dfrac{dr}{dt}\\\Rightarrow \dfrac{dr}{dt} = 0.0064\ ft/sec$

$\dfrac{dA}{dt} = 8\times \pi \times r \dfrac{dr}{dt}\\\Rightarrow \dfrac{dA}{dt} = 8\times 3.14 \times 10 \times 0.0064\\\Rightarrow \dfrac{dA}{dt} = \bold{1.61 \ ft^2/sec}$

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2 years ago