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tatiyna
3 years ago
11

HELP PlZ ASAP NO LINKS

Mathematics
1 answer:
il63 [147K]3 years ago
8 0
0 and 2 is the most logical answer
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The domain for f(x) and g(x) is the set of all real numbers.
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Answer:

Option B

Step-by-step explanation:

=> f(x) = 3x+5

=> g(x) = x^2

Subtracting both

=> (f-g)(x) = 3x+5-x^2

Assembling

=> (f-g)(x) = -x^2+3x+5

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What is the radius of a sphere if its surface area is 2,122.64 square inches? (Use 3.14 for π.)
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3 years ago
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A computer purchased for $1,050 loses 19% of its value every year.
ryzh [129]

Answer:

A) a = 1050 and b = 0.81

B) 3.3

Step-by-step explanation:

Original price of the computer = $ 1050

Rate of decrease in price = r = 19%

This means, every year the price of the computer will be 19% lesser than the previous year. In other words we can say that after a year, the price of the computer will be 81% of the price of the previous year.

Part A)

The exponential model is:

v(t)=a(b)^{t}

Here, a indicates the original price of the computer i.e. the price at time t = 0. So for the given case the value of a will be 1050

b represents the multiplicative rate of change i.e. the percentage that would be multiplied to the price of previous year to get the new price. For this case b would be 81% or 0.81

So, a = 1050 and b = 0.81

The exponential model would be:

v(t)=1050(0.81)^{t}

Part B)

We have to find after how many years, the worth of the computer will be reduced to half. This means we have the value of v which is 1050/2 = $ 525

Using the exponential model, we get:

525=1050(0.81)^{t}\\\\ 0.5=(0.81)^{t}\\

Taking log of both sides:

log(0.5)=log(0.81)^{t}\\\\ log(0.5)=t \times log(0.81)\\\\ t = \frac{log(0.5)}{log(0.81)}\\\\ t = 3.3

Thus, after 3.3 years the worth of computer will be half of its original price.

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3 years ago
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Law Incorporation [45]

Answer:

I really hope this helped, this sentenced seemed the most logical to me.

Step-by-step explanation:

A study at Carnegie University showed that just listening to someone else talk on a phone cause a 37 percent drop in concentration among drivers.

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