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fredd [130]
3 years ago
14

Rachel is a stunt driver. One time, during a gig where she escaped from a building about to explode(!), she drove to get to the

safe zone at 24 meters per second. After 4 seconds of driving, she was 70 meters away from the safe zone
Mathematics
1 answer:
lara31 [8.8K]3 years ago
7 0

Answer:

D(t)=-24t+166

Step-by-step explanation:

Let

D(t) = the distance to the safe zone (measured in meters)

t = time (measured in seconds)

<u>Given:</u>

Rachel's rate = 24 meters per second

At t_1=4 seconds D(t_1)=70 meters

<u>Find:</u> D(t)

Rachel's rate is the slope of the function D(t). Since the distance is decreasing when the time is increasing, the slope must be negative.

Hence, the function expression is

D(t)=-24t+b

To find b, substitute t_1=4 and  D(t_1)=70:

70=-24\cdot 4+b\\ \\70=-96+b\\ \\b=70+96\\ \\b=166

So,

D(t)=-24t+166

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How many times greater is the number 4000 than the number 10?
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The number 4000 is 400 times greater than the the number 10.
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The area of a playground is 336 yd2. The width of the playground is 5 yd longer than its length. Find the length and width of th
kondaur [170]
L=16
W=21

Set up a systems of equations: 
x=length 
y=width 

xy=336
x+5=y

Use substitution to solve: 
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x^2+5x=336 

Solve using factoring: 
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(x-16)(x+21)=0
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. The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to nd
Blababa [14]

The question is:

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation

x²y'' - 7xy' + 16y = 0; y1 = x^4

Answer:

The second solution y2 is

A(x^4)lnx

Step-by-step explanation:

Given the homogeneous differential equation

x²y'' - 7xy' + 16y = 0

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We need to find a second solution y2 using the method of reduction of order.

Let y2 = uy1

=> y2 = ux^4

Since y2 is also a solution to the differential equation, it also satisfies it.

Differentiate y2 twice in succession with respect to x, to obtain y2' and y2'' and substitute the resulting values into the original differential equation.

y2' = u'. x^4 + u. 4x³

y2'' = u''. x^4 + u'. 4x³ + u'. 4x³ + u. 12x²

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Let w = u'

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Integrating both sides

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w = Ae^(-lnx) (where A = e^C)

w = A/x

But w = u'

So,

u' = A/x

Integrating this

u = Alnx

Since

y2 = uy1

We have

y2 = (Alnx)x^4 = (Ax^4)lnx

Therefore, the second solution y2 is

A(x^4)lnx

6 0
3 years ago
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