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3 years ago

The choices are the same for both.

Mathematics

2 answers:

densk [106]3 years ago

5 0

Answer:

f(-∞) -∞

f(∞) = -∞

Step-by-step explanation:

f(x) = -8x^4 +3x^2 -7x

-----------------------

4x^2 -4

The -8x^4 / 4x^2 will dominate at infinity

f(x) can be approximated at ±∞ by -2 x^2

f(-∞) = -2(-∞)^2 = -2 *∞ = -∞

f(∞) = -2(∞)^2 = -2 *∞ = -∞

dlinn [17]3 years ago

3 0

Answer:

- $\infty$

Step-by-step explanation:

f(x)=\frac{-8x^4+3x^2-7x}{4x^2-4}=\frac{-8x^2+3-7/x}{4-4/x^2}

As x approaches -\infty, f(x) will tend to -\infty

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A house cost $120,000 when it was purchased. The value of the house increases by 10% each year. Find the rate of growth each mon

olga_2 [115]

FIRST MODEL:

Well the model for the value of the house is:

$V={ \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000$

V = Value

t = Years passed {t≥0}

-----------------------

When t=0, V=120000

When t=1, V=132000

When t=2, V=145200

etc... etc...

---------------------------

Now, this model is actually curved so there is no constant rate of growth each month. We can only calculate what the rate of growth is at a particular time. If we want to find out the rate of growth at a particular time, we must differentiate the formula (model) above.

--------------------------

$V={ \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000 \right) } }$

$\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ t } \right) } } +\ln { \left( 120000 \right) } \\ \\ \ln { V=t\ln { \left( \frac { 11 }{ 10 } \right) } } +\ln { \left( 120000 \right) }$

$\\ \\ \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \\ \\ V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\\ \\ \therefore \quad \frac { dV }{ dt } =\ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^{ t }\cdot 120000$

Plug any value of (t) that is greater than 0 into the formula above to find out how quickly the investment is growing. If you want to find out how quickly the investment was growing after 1 month had passed, transform t into 1/12.

The rate of growth is being measured in years, not months. So when t=1/12, the rate of growth turns out to be 11528.42 per annum.

SECOND MODEL (What you are ultimately looking for):

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

V = Value of house

t = months that have gone by {t≥0}

Formula above differentiated:

$V={ \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000\\ \\ \ln { V } =\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000 \right) }$

$\\ \\ \ln { V=\ln { \left( { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } } \right) } } +\ln { \left( 120000 \right) }$

$\\ \\ \ln { V=\frac { t }{ 12 } } \ln { \left( \frac { 11 }{ 10 } \right) } +\ln { \left( 120000 \right) }$

$\\ \\ \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) }$

$\\ \\ V\cdot \frac { 1 }{ V } \cdot \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot V$

$\\ \\ \therefore \quad \frac { dV }{ dt } =\frac { 1 }{ 12 } \ln { \left( \frac { 11 }{ 10 } \right) } \cdot { \left( \frac { 11 }{ 10 } \right) }^{ \frac { t }{ 12 } }\cdot 120000$

When t=1, dV/dt = 960.70 (2dp)

dV/dt in this case will measure the rate of growth monthly. As more money is accumulated, this rate of growth will rise. The rate of growth is constantly increasing as the graph of V is actually a curve. You can only find out the rate at which the house value is growing monthly at a particular time.

6 0

3 years ago

Mr brown bought a lot that ia half as long and twice as wide Mrs.Rockwell's lot.How does the area of his lot compare to the area

prohojiy [21]

Answer:

Area of Mrs. Rockwell's lot is equal to the area of Mr. Brown's lot

Step-by-step explanation:

We can suppose the dimensions of Mrs. Rockwell's lot to be:

Length = x

Width = y

Then, we can write the dimensions of Mr. Brown's lot as:

Length = half as long as Mrs. Rockwell's lot

Length = 0.5x

Width = twice as wide as Mrs. Rockwell's lot

Width = 2y

Area of Mrs. Rockwell's lot = Length * Width

= x*y

Area of Mrs. Rockwell's lot = xy

Area of Mr. Brown's lot = 0.5x*2y

Area of Mr. Brown's lot = xy

<u>Area of Mrs. Rockwell's lot </u><u>is equal</u><u> to the area of Mr. Brown's lot, as calculated above</u>.

7 0

4 years ago

You can exchange 25 U.S. Dollars for 20 Euros. What is the exchange rate in U.S. Dollars per Euro?

Leokris [45]

Disclaimer: Exchange rates vary almost on a constant basis, so it is best to always have an up-to-date chart from an online source. For the purposes of this problem, we don't have to worry about the rates changing.

------------------

25 U.S Dollars = 20 Euros

25/20 U.S. Dollars = 20/20 Euros

1.25 U.S. Dollars = 1 Euro

<h3>The exchange rate is 1.25 U.S. Dollars per Euro</h3>

Saying "per Euro" is the same as saying "per 1 Euro". I divided both sides of the second equation by 20 so that I could turn the "20 Euros" into "1 Euro".

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3 years ago

How do you convert 1.71 kg to cg?

Vesnalui [34]

the answer is 171000 cg

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3 years ago

I need help to figure this out I don't understand it

vladimir1956 [14]

Given the cost of $28 to rent a trailer for an x amount of hours:

The expression is 28x

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