Discrete or continuous

22. Use your model to determine when the volume of oil will get down to 500 gallons.

Helen [10]

Answer:

21. y = 75000·0.935^t

22. after 74.6 days

23. y = 27.8112·1.18832^t

24. 18.8% per month

25. 1748

Step-by-step explanation:

22. It is convenient to use the graphing calculator to solve this problem. The number of days is where the exponential curve has the value 500. It is about 74.55 days. (see the first attachment)

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23. y = 27.8112·1.18832^t (see the second attachment)

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24. The rate of change is the difference between the base of the exponential and 1, often expressed as a percentage. The time period is the units of t.

(1.18832 -1) × 100% ≈ 18.8% . . . . per month

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25. Evaluating the function for t=24 gives y ≈ 1748.30425259 ≈ 1748.

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<em>Comment on graphing calculator</em>

A graphing calculator can make very short work of problems like these. It is worthwhile to get to know how to use one well.

8 0

3 years ago

A determined gardener has 120ft of deer-resistant fence. She wants to enclose a rectangular vegetable garden in her backyard, an

Pavlova-9 [17]

Answer:

the answer is 123

Step-by-step explanation:

cuz thats wat it is cux i think so

3 0

3 years ago

Find the volume for the following. Show your work and include the units.<br> 3 cm<br> 12 cm

Gemiola [76]

Answer:

12 CM is the best answer.

please thanks and follow me

7 0

3 years ago

Read 2 more answers

Find the integral using substitution or a formula.

Nadusha1986 [10]

$\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx$

Derivative of the denominator:
$\rm (x^2+2x+5)'=2x+2$

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

$\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx$

and then split the fraction,

$\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx$

Based on our previous test, we know that a simple substitution will work for the first integral: $\rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx$

So the first integral changes,

$\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx$

integrating to a log,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx$

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

$\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2$

So we have this going on,

$\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx$

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx$

Bringing all that stuff together as a single square,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx$

Making the substitution: $\rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx$

giving us,

$\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du$

simplying a lil bit,

$\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du$

and hopefully from this point you recognize your arctangent integral,

$\rm ln|x^2+2x+5|+\frac52arctan(u)$

undo your substitution as a final step,
and include a constant of integration,

$\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c$

Hope that helps!
Lemme know if any steps were too confusing.

8 0

3 years ago

Look at the photo, please.

ruslelena [56]

Answer:

×=6

good luck!!

4 0

3 years ago

Read 2 more answers

Discrete or continuous ​

Discrete or continuous