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2 years ago

3. A drone takes off from the base of a mountain at a height of 27 ft below sea level. The drone

Mathematics

1 answer:

Korolek [52]2 years ago

5 0

Answer:

669ft high

Step-by-step explanation:

642+27=669ft

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The table shows the different packs of sports cards available for purchase. What is the smallest number of baseball and football

WARRIOR [948]

<span>The least common multiple would be 280 cards of each.

20 Packs of football, 8 packs of Hockey, and 7 packs of Baseball.</span>

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2 years ago

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What are the domain and range of f(x) = 2x - 41?

Darina [25.2K]

The domain of the function is a (-∞, 0) and the range of the function will be f(x) < 0. Then the correct option is C.

<h3>What are domain and range?</h3>

The domain means all the possible values of x and the range means all the possible values of y.

The function is given below.

f(x) = 2x – 41

Then the domain of the function is a (-∞, 0) and the range of the function will be f(x) < 0.

Then the correct option is C.

More about the domain and range link is given below.

brainly.com/question/12208715

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1 year ago

Which number is a composite number? 11 21 31 41

erastovalidia [21]

Answer:

Step-by-step explanation:

It is 21 since 21 = 3 * 7.

The other 3 numbers are prime.

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2 years ago

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Question in picture solve

Anna007 [38]

I think it’s 3 times because 2*3=6 and 4*3=12 and 6*3=18

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2 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

2 years ago