A Susan B. Anthony dollar is in the shape of a regular polygon. it has 11 sides.

If a /b = p /q, then (a + b )/b = _______.<br> (p + b )/q<br> (p + p)/q<br> (p + q)/q<br> (p + a)/q

maxonik [38]

Hope this helps you!

5 0

3 years ago

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What value of s in the equation S/6=18?<br>A.108<br>B.24<br>C.12<br>D.3

Fiesta28 [93]

Answer:

108

Step-by-step explanation:

s=18x6

s=108

5 0

3 years ago

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Find two consecutive even integers such that twice the smaller is 16 more than the larger

drek231 [11]

<u>Define x:</u>
Let one of the numbers be x.
The other number is x + 2

<u>Construct equation:</u>
<span>twice the smaller is 16 more than the larger
</span>⇒2x = x + 2 - 16

<u>Solve x:</u>
2x = x + 2 - 16
x = -14

<u>Find the two numbers:</u>
Smaller number = x = -14
Larger number = x + 2 = -12

Answer: The two numbers are -14 and -12

6 0

3 years ago

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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:

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