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grin007 [14]
3 years ago
6

∆ABC is translated 6 units up and 3 units left to create ∆A'B'C'. If vertex A is at (-1, 2) and vertex B is at (1, 5), then vert

ex A' is at
Mathematics
2 answers:
Paladinen [302]3 years ago
4 0
<span>So we want to know the new coordinates for the vertex A'(x,y) if we know that the vertex A is at A(-1,2) and vertex B is at B(1,5) and that the triangle ABC is translated 6 units up and 3 units left. So the method is simply to add units 6 to x and 3 to y of A to get A'. Going left means we need to go to negative x direction and going up means we need to go to positive y direction. So: A'(-1-3,2+6) and that is: A'(-4,8). </span>
sineoko [7]3 years ago
3 0

Answer: (-4,8)

Step-by-step explanation:

The translation rule for translating a point h units left  is given by :-

(x,y)\to (x-h,k)

The translation rule for translating a point k units up is given by :-

(x,y)\to (x,y+k)

Given : ∆ABC is translated 6 units up and 3 units left to create ∆A'B'C'. If vertex A is at (-1, 2) and vertex B is at (1, 5).

Then, the vertex A'  will be :-

A(-1,2)\to(-1-3,\ 2+6)=(-4,\ 8)

Hence, the vertex A' is at (-4,8).

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Jackie made $28 babysitting last week. her brother joe made only 86% as much as she did. how much did joe make?
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4 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.

<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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