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stich3 [128]
3 years ago
9

Select the correct answer from each drop-down menu.

Mathematics
1 answer:
kifflom [539]3 years ago
3 0

Answer:

  1. 0.83
  2. 0.9
  3. steeper than
  4. slower than

Step-by-step explanation:

Letting t=1 in Ted's equation, we find that he climbs 5/6 stairs in 1 second. As a decimal, 5/6 ≈ 0.83.

Michael climbs 9 stairs in 10 seconds so his rate is ...

... (9 stairs)/(10 seconds) = (9/10) stairs/second = 0.9 stairs/second

Michael's graph will be a line with a slope of 0.9; Ted's graph will be a line with a slope of about 0.83, so the line on Michael's graph is steeper.

Ted climbs fewer stairs per second, so his rate is slower than Michael's.

_____

<em>Comment on the problem</em>

You're being asked to compare two different rates that are associated with two different people. First the comparison is one way, then it is the other way. This can be confusing. It might be helpful to draw and label a simple chart to help you keep it straight. (The attachment is such a chart scribbled on a bit of scratch paper. It is sufficient for the purpose.)

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(a) By inspection, find a particular solution of y'' + 2y = 14. yp(x) = (b) By inspection, find a particular solution of y'' + 2
SOVA2 [1]

Answer:

(a) The particular solution, y_p is 7

(b) y_p is -4x

(c) y_p is -4x + 7

(d) y_p is 8x + (7/2)

Step-by-step explanation:

To find a particular solution to a differential equation by inspection - is to assume a trial function that looks like the nonhomogeneous part of the differential equation.

(a) Given y'' + 2y = 14.

Because the nonhomogeneus part of the differential equation, 14 is a constant, our trial function will be a constant too.

Let A be our trial function:

We need our trial differential equation y''_p + 2y_p = 14

Now, we differentiate y_p = A twice, to obtain y'_p and y''_p that will be substituted into the differential equation.

y'_p = 0

y''_p = 0

Substitution into the trial differential equation, we have.

0 + 2A = 14

A = 6/2 = 7

Therefore, the particular solution, y_p = A is 7

(b) y'' + 2y = −8x

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x

2Ax + 2B = -8x

By inspection,

2B = 0 => B = 0

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x

(c) y'' + 2y = −8x + 14

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = -8x + 14

2Ax + 2B = -8x + 14

By inspection,

2B = 14 => B = 14/2 = 7

2A = -8 => A = -8/2 = -4

The particular solution y_p = Ax + B

is -4x + 7

(d) Find a particular solution of y'' + 2y = 16x + 7

Let y_p = Ax + B

y'_p = A

y''_p = 0

0 + 2(Ax + B) = 16x + 7

2Ax + 2B = 16x + 7

By inspection,

2B = 7 => B = 7/2

2A = 16 => A = 16/2 = 8

The particular solution y_p = Ax + B

is 8x + (7/2)

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