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ipn [44]
3 years ago
9

Two mechanics worked on a car. The first mechanic charged $95  per hour, and the second mechanic charged $60  per hour. The mech

anics worked for a combined total of 20  hours, and together they charged a total of $1375 . How long did each mechanic work?
Mathematics
1 answer:
Yanka [14]3 years ago
3 0

Answer:

<em>Mechanic A worked for 5 hours</em> and <em>Mechanic B worked for 15 hours</em>

I hope my answer and explanation helped!

<u>okay to get started you need to make a system of equations:</u>

x= number of hours worked by mechanic A

y= number of hours worked by mechanic B

x + y= 20

95x + 60y= 1375

<u>substitute in an equation:</u>

x + y= 20

y= 20- x

95x + 60(20-x)=1375

<u>Solve for x</u>

95x + 1200 - 60x=1375

35x =175

x= 5

<u>plug in x to solve for y</u>

x + y= 20

5 + y= 20

y=15

<u>Check work</u>

then you're done :D

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tresset_1 [31]

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Read 2 more answers
Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1
belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

            F(x) = \int\limits^a_b {f(x)} \, dx

By using integration formula

\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c

Given  x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

 \int\limits^6_^-6} (x^{5}  + 8 x^{4}  + 2 x^{2}  + 5 x + 15) )dx

<em>On integration , we get</em>

=   (\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}

F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)

= (\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))

After simplification and cancellation we get

 =  \frac{2 X 8 X (6)^{5} }{5} + \frac{2 X 2 X (6)^3}{3} + 2 X 15 X 6

on calculation , we get

= \frac{124,416}{5} + \frac{864}{3} + 180

On L.C.M  15

= \frac{124,416 X 3 + 864 X 5 + 180 X 15}{15}

= 25 351.2 units²

<u><em>Conclusion</em></u>:-

<em>The area of the region between the graph of the given function and the x-axis = 25,351 units²</em>

6 0
3 years ago
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