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klasskru [66]
3 years ago
15

Gina and Lucy go to the library at 3:30 p.m. They need to be at home at 4:45 p.m. It takes them 15 minutes to walk to the librar

y. How many minutes can they spend at the library?
Mathematics
1 answer:
MrRa [10]3 years ago
7 0
4:45 -3:30 -2*(0:15) = 0:45

Gina and Lucy can spend 45 minutes at the library.

_____
They will get to the library at 3:45. They must start home by 4:30. From 3:45 to 4:00 is 15 minutes, and it is 30 more minutes to 4:30. The time they can spend at the library totals 45 minutes.
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Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

The price p, in dollars, of a specific car that is x year old is modeled by the function p(x)=22,255(0.91)^x

a) to determine the cost of a 2 year old car, we will substitute 2 for x in the given function. Therefore

p(2)=22,255(0.91)^2

p(2)=22,255 × 0.8281 = $18673.655

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b) to determine the cost of a 7 year old car, we will substitute 7 for x in the given function. Therefore

p(7)=22,255(0.91)^7

p(2)=22,255 × 0.51676101936 = 11500.51648579693

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c) 0.91 indicates exponential decay rate. It is a fixed percentage by which the value of the car decreases every year. It is determined by (1 - rate of decay)

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bezimeni [28]

The question can't be solved because we can't have negative sign under square root

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-2x + y = 5 <br><br> what does y equal and how did you solve it?
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Answer:

y=2x+5

Step-by-step explanation:

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You find the eigenvalues of a matrix A by following these steps:

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So, in this case, we have

A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]

The determinant of this matrix is

\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4

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