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elena-s [515]
3 years ago
5

a spherical balloon is being inflated so that the radius is increasing at a rate of 3 cm/sec. find the rate at which the volume

is increasing when r=10cm.
Mathematics
1 answer:
neonofarm [45]3 years ago
3 0

Answer:  1200pi cubic cm per sec

========================================

Work Shown:

dr/dt = 3 is the rate of change of the radius

V = (4/3)pi*r^3 ..... volume of a sphere

dV/dt = (4/3)*3*pi*r^2*dr/dt ... chain rule

dV/dt = 4*pi*r^2*dr/dt

dV/dt = 4*pi*r^2*3

dV/dt = 12*pi*r^2

dV/dt = 12*pi*10^2 ... plug in r = 10

dV/dt = 1200pi

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(1 point) For the function f graphed below, find the following limits:
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1. ∞

2. -∞

3. ∞

4. 1

5. -∞

<h3>What is a limit?</h3>

A limit is given by the <u>value of function f(x) as x tends to a value</u>.

For this problem, at x = 0, we have that to the left the function goes to positive infinity, while to the right it goes to negative infinity, hence:

1. lim f(x) = ∞

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2. lim f(x) = -∞

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At x = 2, the function goes to infinity to the left and to the right, hence:

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To the left of the graph, the function goes to negative infinity, while to the right it goes to 1, hence:

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7 0
2 years ago
A rectangular field is 75 yards wide and 105 yards long
nika2105 [10]

Answer:

width = 72 yards

length = 108 yards

Step-by-step explanation:

Given:

  • Width = 75 yards
  • Length = 105 yards

<u>Area of the field</u> with the given values:

\begin{aligned}\textsf{Area of a rectangle}&=\sf width \times length\\& = \sf 75 \times 105\\& = \sf 7875\:\: yd^2\end{aligned}

To maintain the <u>same perimeter</u>, but <u>change the area</u>, either:

  • decrease the width and increase the length by the same amount, or
  • increase the width and decrease the length by the same amount.

In geometry, length pertains to the <u>longest side</u> of the rectangle while width is the <u>shorter side</u>.  Therefore, we should choose:

  • decrease the <u>width</u> and increase the <u>length</u> by the <u>same amount</u>.

<u>Define the variables</u>:

  • Let x = the amount by which to decrease/increase the width and length.

Therefore:

\implies \sf width \times length < 7875\:\:yd^2

\implies (75-x)(105+x) < 7875

Solve the inequality:

\begin{aligned}(75-x)(105+x) & < 7875\\7875-30x-x^2 & < 7875\\-x^2-30x & < 0\\-x(x+30) & < 0\\x(x+30) & > 0\\\implies x & > 0 \:\: \textsf{ or }\:\:x < - 30\end{aligned}

Therefore, as distance is positive only and the maximum width is 75 yd (since we are subtracting from the original width):

\begin{cases}\textsf{width} = 75 - x\\\textsf{length} = 105 + x\end{cases}

\textsf{where } 0 < x < 75

Therefore, to find the width and length of another rectangular field that has the same perimeter but a smaller area than the first field, simply substitute a value of x from the restricted interval into the found expressions for width and length:

<u>Example 1</u>:

  • Let x = 3

⇒ Width = 75 - 3 = 72 yd

⇒ Length = 105 + 3 = 108 yd

⇒ Perimeter = 2(72 + 108) = 360 yd

⇒ Area = 72 × 108 = 7776 yd²

<u>Example 2</u>:

  • Let x = 74

⇒ Width = 75 - 74 = 1 yd

⇒ Length = 105 + 74 = 179 yd

⇒ Perimeter = 2(1 + 179) = 360 yd

⇒ Area = 1 × 179 = 179 yd²

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