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USPshnik [31]
3 years ago
11

Suppose that a car rental agency charges a fixed amount per day plus an amount per mile for renting a car. Heidi rented a car on

e day and paid $52 for 180 miles. On another day she rented a car from the same agency and paid $77.50 for 350 miles. Find the linear function f(x) (where x is in miles and f(x) is in dollars) that the agency could use to determine its daily rental charges.
f(x) =
Mathematics
1 answer:
elena55 [62]3 years ago
8 0

9514 1404 393

Answer:

  f(x) = 0.15x +25

Step-by-step explanation:

We are given two points:

  (x, y) = (180, 52) or (350, 77.50)

The slope is given by ...

  m = (y2 -y1)/(x2 -x1)

  m = (77.50 -52)/(350 -180) = 25.50/170 = 0.15

The y-intercept is given by ...

  b = y -mx

  b = 52 -0.15×180 = 25

Then the equation for the linear function is ...

  f(x) = mx +b

  f(x) = 0.15x +25

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fredd [130]
Data:

Formula: diameter = 2*radius

Formula: (<span>Circle length)
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<span>
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3 years ago
Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

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Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

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Step-by-step explanation:

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