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Svetach [21]
3 years ago
5

Allison claims that the (triangle)ABQ is similar to (triangle)RPQ, given that AB and PR are parallel. Which of Allison's claims

supporting her argument are correct? Select all that apply. <1 = <2 because they are vertically opposite angles.<ABQ = <QPR because they are corresponding angles.<1 = <2 because they are alternate interior angles.<BAQ = <QRP because they are alternate interior angles.(triangle)ABQ and (triangle)RPQ are not similar by AA similarity.

Mathematics
1 answer:
S_A_V [24]3 years ago
3 0
I draw the two triangles, see the picture attached.

As you can see, angle 1 and 2 are vertically opposite angles because they are formed by the same two crossing lines and they face each other.

Angles <span>ABQ and QPR, as well as angles BAQ and QRP, are alternate interior angles because they are formed by </span><span>two parallel lines crossed by a transversal, and they are inside the two lines on opposite sides of the transversal.</span>

Hence, Allison's correct claims are:
1 = 2 because they are vertically opposite angles. 
BAQ = QRP because they are alternate interior angles. 

Therefore Allison, in order to prove her claim, can use the AA similarity theorem: if two angles of a triangle are congruent to two angles of the other triangle, then the two triangles are similar. 

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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

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Reverting the change of variable, we have:

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

Have a nice day!

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