2x+2y =50
2(7)+ 2(3)=50
14+6 =50
20=/=50

This is false because when you substitute the variables with their numbers and then multiply and add, your answer does not equal 50

5 0

4 years ago

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A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 3 centimeters. Assuming the balloon i

Anni [7]

Answer:

The radius is growing at the rate of 0.09 centimeter per second at the instant it pops.

Step-by-step explanation:

We are given the following information in the question:

A spherical balloon is designed to pop at the instant its radius has reached 3 centimeters.

The balloon is filled with helium at a rate of 10 cubic centimeters per second .

$\displaystyle\frac{dV}{dt} = 10~cm^3/sec.$

We have to find fast the radius is growing at the instant it pops.

Volume of spherical balloons =

$\displaystyle\frac{4}{3}\pi r^3$

Differentiating, we get,

$\displaystyle\frac{dV}{dt} = \frac{4}{3}\pi 3r^2 \displaystyle\frac{dr}{dt} = 4\pi r^2 \displaystyle\frac{dr}{dt}$

Putting the values, we get,

$10 = 4\times 3.14\times (3)^2\times \displaystyle\frac{dr}{dt}\\\\\frac{dr}{dt} = \frac{10}{4\times 3.14\times (3)^2} = 0.0884 \approx 0.09\\\\\frac{dr}{dt} = 0.09\text{ centimeter per second}$

The radius is growing at the rate of 0.09 centimeter per second at the instant it pops.

4 0

3 years ago