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Nezavi [6.7K]
3 years ago
6

Which graph represents the function f (x) = StartFraction 1 Over x EndFraction minus 1?

Mathematics
1 answer:
Arturiano [62]3 years ago
6 0

Answer:

The last answer, D

Step-by-step explanation:

i got it right on ed

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How much would $600 invested at 8% interest compounded annually be worth after 3 years? Round your answer to the nearest cent.
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Answer:

Step-by-step explanation:

A = P (1 + i)³

A = final amount

P = principal

i = interest

3 = 3 years

A = $600 (1 + 0.08)³

A = $600 (1.26)

A = $756

5 0
3 years ago
Find the area of the shaded region. Round your answer to the nearest tenth.
Dmitrij [34]

Answer:

10.5cm^2

Step-by-step explanation:

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3 years ago
How do you find the radius when given the volume and height? ​
Eduardwww [97]

Answer:

Calculate the area of the base (which is a circle) by using the equation πr² where r is the radius of the circle. Then, multiply the area of the base by the height of the cylinder to find the volume. SOO just remember to divide the diameter by two to get the radius. If you were asked to find the radius instead of the diameter, you would simply divide 7 feet by 2 because the radius is one-half the measure of the diameter. The radius of the circle is 3.5 feet. You can also use the circumference and radius equation.

Step-by-step explanation:

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4 0
3 years ago
Question 3
tino4ka555 [31]
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5 0
2 years ago
A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its
VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate (\dot V), measured in liters per hour, is directly proportional to draining time (t), measured in hours. That is:

\dot V \propto \frac{1}{t}

\dot V = \frac{k}{t} (Eq. 1)

Where k is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) <em>Large and small drains are opened</em>

\dot V_{s}+\dot V_{l} = \frac{k}{2} (Eq. 2)

\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}

(ii) <em>Only the small drain is opened</em>

\dot V_{s} = \frac{k}{t_{l}+3} (Eq. 3)

\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}

(iii) <em>Only the big drain is opened</em>

\dot V_{l} = \frac{k}{t_{l}} (Eq. 4)

\frac{\dot V_{l}}{k}  = \frac{1}{t_{l}}

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}

\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}

2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}

t_{l}^{2}-t_{l}-3 = 0 (Eq. 5)

Whose roots are determined by the Quadratic Formula:

t_{l,1}\approx 2.303\,h and t_{l,2} \approx -1.302\,h

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

t_{s} = 2.303\,h+3\,h

t_{s} = 5.303\,h

The draining time when only the small drain is opened is 5.303 hours.

7 0
3 years ago
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