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Vikki [24]
3 years ago
9

I need some help here

Mathematics
1 answer:
grandymaker [24]3 years ago
3 0
In 1 and 2, you are asked to find "a" and "b" in the pattern
  g(x) = a•bˣ

1: selection 2
2: selection 4

3. You want the result of evaluating 4000*(1.06)⁵ ≈ 5352.90, selection 2.
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When calculating compound interest, the periodic interest rate is always less than the annual interest rate. True or false
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Step-by-step explanation: If your Compounding interest, the periodic interest is Sometimes less than the annual interest because the period of interest could be say 2 years

Hope this helps ☺  


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Choose yes or no to tell is the fraction 1/2 will make each equation true. 1/2 + 5/5 = 3/2​
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Answer:

yes ?-?

Step-by-step explanation:

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Why is 743 a prime number
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743 is a prime number because it has no other positive divisors besides one and itself.
7 0
3 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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