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chubhunter [2.5K]
3 years ago
15

The original price of a car is 18,000, and it depreciates by 15% each year. What is the value of the car after three years?

Mathematics
1 answer:
Rashid [163]3 years ago
3 0

Answer: $9,900


Step-by-step explanation:

Convert the percentage into a decimal: 0.15

Multiply 0.15 by 18000

The product is 2700

Multiply 2700 by the 3 years

The product is 8100

Subtract 8100 from 18000

The result is $9,900

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Find three consecutive integers such that the sum of twice the smallest and 3 times the largest is 126
True [87]

Answer:

Step-by-step explanation:

let the numbers be x,x+1,x+2   (x<x+1<x+2)

2x+3(x+2)=126

2x+3x+6=126

5x=126-6

5x=120

x=120/5=24

numbers are 24,24+1,24+2

or 24,25,26

5 0
2 years ago
Write the quadratic function in standard form.<br><br> y=2(x - 3)² +9
xz_007 [3.2K]

Answer:

y = 2x^2 - 12x + 27

Step-by-step explanation:

<u>Step 1:  Distribute the power</u>

y = 2(x - 3)² + 9

y = 2(x^2 - 6x + 9) + 9

y = 2x^2 - 12x + 18 + 9

y = 2x^2 - 12x + 27

Answer: y = 2x^2 - 12x + 27

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3 years ago
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IrinaK [193]

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6 0
3 years ago
A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria
Kaylis [27]

Answer:

A) The expression for the number of bacteria is P(t) = 400e^{0.7783t}.

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

P' = kP

where P stands for the population of bacteria.

Writing P' as \frac{dP}{dt}, we get

\frac{dP}{dt} = kP.

Notice that this is a separable equation, so

\frac{dP}{P} = kdt.

Then, integrating in both sides of the equality:

\int\frac{dP}{P} = \int kdt.

We have,

\ln P = kt+C.

Now, taking exponential

P(t) = Ce^{kt}.

The next step is to find the value for the constant C. We do this using the initial condition P(0)=400. Recall that this is the initial population of bacteria. So,

400 = P(0) = Ce^{k0}=C.

Hence, the expression becomes

P(t) = 400e^{kt}.

Now, we find the value for k. We are going to use that P(4)=9000. Notice that

9000 = 400e^{k4}.

Then,

\frac{90}{4} = e^{4k}.

Taking logarithm

\ln\frac{90}{4} = 4k, so \frac{1}{4}\ln\frac{90}{4} = k.

So, k=0.7783788273, and approximating to the fourth decimal place we can take k=0.7783. Hence,

P(t) = 400e^{0.7783t}.

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

P(5) =400e^{0.7783*5} = 19593.723 \approx 19593.  

C) In this case we want to do the reverse operation: we want to find the value of t such that

30000 = 400e^{0.7783t}.

This expression is equivalent to

75 = e^{0.7783t}.

Now, taking logarithm we have

\ln 75 = 0.7783t.

Finally,

t = \frac{\ln 75}{0.7783} \approx 5.55.

So, after 5.55 hours the population of bacteria will reach 30000.

6 0
3 years ago
Morgan has $60 to spend on clothes. He wants to buy a pair of jeans for $25 and spend the rest on t-shirts. Each t-shirt costs $
irakobra [83]
1 pair of jeans to 7 shirts
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