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vlada-n [284]
2 years ago
10

Jake fills a tank that can hold 200 gallons of water. The tank already has 50 gallons of water in it when Jake starts filling it

at the rate of 10 gallons per minute. Karla fills a tank that can hold 300 gallons of water. That tank already has 100 gallons of water in it when Karla starts filling it at the rate of 5 gallons per minute. Jake and Karla start filling the tanks at the same time. How long after they start filling the tanks do the tanks have the same volume of water? What is that volume of water? I​
Mathematics
1 answer:
Ede4ka [16]2 years ago
4 0

Answer:

Let m represent the number of minutes.

Amount of water in Karla's tank:  100 + 5m

Amount of water in Jake's tank:  50 + 10m

Same amount of water:  100 + 5m  =  50 + 10 m

--->                                         100  =  50 + 5m

--->                                            50  =  5m

--->                                            10 = m

In ten minutes, Karla's tank will have 100 + 5(10)  =  100 + 50  =  150 gallons.

In ten minutes, Jake's tank will have 50 + 10(10)  =  50 + 100  =  150 gallons.

Step-by-step explanation:

hope this helped!

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<h2>Answer</h2>

After the dilation \frac{5}{3} around the center of dilation (2, -2), our triangle will have coordinates:

R'=(2,3)

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<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

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Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

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R=(2,1)

R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})

R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})

R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})

R'=(2,3)

S=(2,-2)

S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})

S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

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T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})

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Now we can finally draw our triangle:

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