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vazorg [7]
3 years ago
8

Chase collects $27.34 for a fundraiser Sydney collects$9.83 more than Chase ally collects 4times as much as Sydney. how much doe

s ally collects. please explain by step by step guide me to
Mathematics
2 answers:
adelina 88 [10]3 years ago
8 0
So Chase has $27.34.
You add $27.34 to $9.83. (Which is $37.17).
That will give you the amount of how much Sydney has.
To find how much Ally has you times $37.17 by 4.
Basically, you are doing $37.17 + $37.17 + $37.17 + $37.17.
When you have finished adding them you will get the answer.
So the answer is $148.68
grigory [225]3 years ago
8 0
Chase collects $27.34 and Sydney got $9.83 more. So you add $27.34 and 9.83 given Sydney a total of $37.17. Therefore if Ally collected 4 times more than Sydney you will multiply $37.17 by 4 given Ally a total of $148.68
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Convert to smaller parts if possible​
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Answer:

Row 1 -

1/3, 1/2, 1/3, 2/3, 2/3, 8/15, 1/2.

Row 2 -

3/4, 3/4, 2/7, 21/25, 5/6, 7/9, 1/3.

Row 3 -

3/20, 7/20, 3/25, 3/5, 3/5, 1, 3/2 OR 1 1/2.

Hope this helped you out.

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During a bike challenge,riders have to collect various colored ribbons. Each 1/2 mile they collect a red ribbon, each 1/8 mile t
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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.

<h2>Why?</h2>

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
3 years ago
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