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levacccp [35]
3 years ago
13

Two mechanics worked on a car. The first mechanic worked for 20 hours, and the second mechanic worked for 5 hours. Together they

charged a total of 2575. What was the rate charged per hour by each mechanic if the sum of the two rates was 170$ per hour?
Mathematics
2 answers:
fiasKO [112]3 years ago
8 0

Answer:

The rate charged by the first mechanic is $115 and the second mechanic is $55.

Step-by-step explanation:

Let, the rate charged by the first mechanic = $x and the rate charged by the second mechanic = $y.

Since, the first mechanic worked for 20 hours and the second mechanic worked for 5 hours.

Thus, the total rate charged by both the mechanics is (20x+5y).

As, they were charged total $2575.

So, we get, 20x+5y = 2575.

Moreover, the sum of the rate is given to $170.

This gives us, x + y = 170

So, we get the system of equations as,

20x+5y = 2575 i.e. 4x + y = 515

x + y = 170

Now, we will solve the equations by subtracting them both,

We get, 3x = 345 i.e. x = 115

So, y = 170 - x i.e. y = 170 - 115 i.e. y = 55

Hence, the rate charged by the first mechanic is $115 and the second mechanic is $55.

SashulF [63]3 years ago
8 0

Answer:

x=115  by first mechanic and y=55 by second mechanic.

Step-by-step explanation:

Let the first mechanic charge be x

And the second mechanic charge  be y

First mechanic worked for 20 hours that means 20x

And second mechanic worked for 5 hours that means 5y

in total they charged for  2575

So, 20 x + 5 y =2575 (1)

And Sum of two rates was:  x+y=170  (2)

Solving above two equations  (1) and (2) we get:

x=115 and y=55.

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(12+x)(14+x)=252 \\ \\ 168+12x+14x+x^2-252=0 \\ \\ x^2+26x-84=0 \\ \\ \\ Using \ quadratic \ formula: \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ a=1 \\ \\ b=26 \\ \\ c=-84 \\ \\ \\ x=\frac{-26 \pm \sqrt{26^2-4(1)(-84)}}{2(1)} \\ \\ x=\frac{-26 \pm \sqrt{1012}}{2} \\ \\ \\ Two \ solutions: \\ \\ x_{1}=-13+\sqrt{253} \approx 2.9\\ \\ x_{2}=-13-\sqrt{253} \approx -28.9 \\ \\ x_{2} \ is \ discarded \ because \ it \ can't \ be \ negatives

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<em>In order to increase the area of a rectangular garden that measures 12 feet by 14 feet by 50% Jake must increase each dimension by equal lengths, x:</em>

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