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Arlecino [84]
2 years ago
14

A freight train completes its journey of 150 miles 1 hour earlier if its original speed is increased by 5 miles/hour what is the

train’s original speed
Mathematics
1 answer:
vlabodo [156]2 years ago
7 0
<span>A freight train completes its journey of 150 miles 1 hour earlier if its original speed is increased by 5 miles/hour. What is the train’s original speed?
***
let x=original speed
x+5=increased speed
travel time=distance/speed
..
lcd:x(x+5)
150(x+5)-150x=x(x+5)
150x+750-150x=x^2+5x
x^2+5x-750=0
(x-25)(x+30)=0
x=25
What is the train’s original speed? 25 mph</span>
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Twice a year, for all four years of college, you used your credit card to pay for books and supplies, which amounts to $516 a se
mixer [17]

Answer:

$11,728

Step-by-step explanation:

Twice a year, for 4 years is 8 times

516×8 = 4128

Yearly for 4 years is 4 times

700×4 = 2800

Every month for 4 years is 48 times

100×48 = 4800

Minimum expenditure:

4128 + 2800 + 4800 = $11,728

4 0
2 years ago
Robert has $50 to spend on his utility bills each month. The basic monthly charge is 23.77 electricity costs 0.1117 for each kil
Lilit [14]

Answer:

<em>The maximum number of kilowatt-hours is 235</em>

Step-by-step explanation:

<u>Inequalities</u>

Robert's monthly utility budget is represented by the inequality:

0.1116x + 23.77 < 50

Where x is the number of kilowatts of electricity used.

We are required to find the maximum number of kilowatts-hours used without going over the monthly budget. Solve the above inequality:

0.1116x + 23.77 < 50

Subtracting 23.77:

0.1116x < 50 - 23.77

0.1116x < 26.23

Dividing by 0.1116:

x < 26.23/0.1116

x < 235

The maximum number of kilowatt-hours is 235

7 0
1 year ago
Mike has $600 in his bank account over the last five years the amount in his account has went down to $250 what integer represen
UkoKoshka [18]

600 ---> 250

600 - 250 = 350

350 / 5 = 70

Average decrease in value per year: $70

7 0
2 years ago
The sides of a square are five to the power of two fifths inches long. What is the area of the square?
Feliz [49]
I believe the correct answer from the choices listed above is the first option. If the  sides of a square are five to the power of two fifths inches long, then the are of the square would be <span>five to the power of four fifths square inches. Hope this answers the question.</span>
3 0
2 years ago
The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

6 0
2 years ago
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