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Vanyuwa [196]
3 years ago
15

Which of the following represents 12% growth per time period?

Mathematics
1 answer:
Andreyy893 years ago
5 0

Answer:

f(x) = 1,000(1.12)x

Step-by-step explanation:

You can use your context clues to figure out this answer.

Your goal is to find an answer that :

1) The number is greater than 1,000 (since it grows)

2) The number doesn't increase too much, but a close estimate to 12%

A percentage (that is less than 100%) is a decimal number. A percentage represents that number out of 100. So, 12% is only 12 out of 100.

Meaning that .12 = 12%. But in this case, we want an increase of 12%. Since we want it to grow from our initial value, <em>1,000</em>, you want to find where the multiplier is greater than 1. Because 100% of 1000 is 1000. + .12 = 112%.

If you put in your calculator, 1000 * 1.12, you should get 1,120

. g(x) actually decreases 1,000. It actually gets 12% of 1,000.

h(x) represents 80% of 1,000 but not an increase of 12%.

k(x) is such a huge number and should automatically be negated/crossed off.

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Angel bought 5 pounds of chicken for $9.99 . what is the unit rate for 1 pound of chicken
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Arlene is comparing the prices of two truck rental companies. Company A charges $7 per hour and an additional $30 as service cha
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Part A: 7n + 30= -5 and 8n + 40= 16

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Part C: Company A saves you 12.

Step-by-step explanation:

Company A: 7n + 30

Company B: 8n + 40

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A tank contains 10 liters of pure water. Saline solution with a variable concentration 5 grams of salt per liter is pumped into
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Answer:

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

Step-by-step explanation:

The mass flow rate dQ(t)/dt = mass flowing in - mass flowing out

Since 5 g/L of salt is pumped in at a rate of 4 L/min, the mass flow in is thus 5 g/L × 4 L/min = 20 g/min.

Let Q(t) be the mass present at any time, t. The concentration at any time ,t is thus Q(t)/volume = Q(t)/10. Since water drains at a rate of 4 L/min, the mass flow out is thus, Q(t)/10 g/L × 4 L/min = 2Q(t)/5 g/min.

So, dQ(t)/dt = mass flowing in - mass flowing out

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Since the salt just begins to be pumped in, the initial mass of salt in the tank is zero. So Q(0) = 0

So, the initial value problem is thus

dQ(t)/dt = 20 - 2Q(t)/5 , Q(0) = 0

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