9514 1404 393
Answer:
a. 0.6 h
b. 1.3 h
c. C -- No
d. 4.2 mi/h
Step-by-step explanation:
a. The relationship between time (t), speed (s), and distance (d) is ...
t = d/s
The time spent running 4 miles at 7 miles per hour is ...
t = (4 mi)/(7 mi/h) = (4/7) h ≈ 0.6 h
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b. The time spent walking home is ...
t = (4 mi)/(3 mi/h) = 4/3 h ≈ 1.3 h
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c. C -- No, less time is spent at 7 mi/h
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d. The average speed is ...
s = d/t = (4 mi +4 mi)/(4/7 h +4/3 h) = (8 mi)/(40/21 h) = 4.2 mi/h
I'm taking a guess that regular price would be $1,106.64 plus the 6% tax
100 times longer because the sand grand is a very tiny particle
Answer:
f) a[n] = -(-2)^n +2^n
g) a[n] = (1/2)((-2)^-n +2^-n)
Step-by-step explanation:
Both of these problems are solved in the same way. The characteristic equation comes from ...
a[n] -k²·a[n-2] = 0
Using a[n] = r^n, we have ...
r^n -k²r^(n-2) = 0
r^(n-2)(r² -k²) = 0
r² -k² = 0
r = ±k
a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q
We find p and q from the initial conditions.
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f) k² = 4, so k = 2.
a[0] = 0 = p + q
a[1] = 4 = -2p +2q
Dividing the second equation by 2 and adding the first, we have ...
2 = 2q
q = 1
p = -1
The solution is a[n] = -(-2)^n +2^n.
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g) k² = 1/4, so k = 1/2.
a[0] = 1 = p + q
a[1] = 0 = -p/2 +q/2
Multiplying the first equation by 1/2 and adding the second, we get ...
1/2 = q
p = 1 -q = 1/2
Using k = 2^-1, we can write the solution as follows.
The solution is a[n] = (1/2)((-2)^-n +2^-n).
Answer:
We conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.
Step-by-step explanation:
We know that the perimeter of a rectangle = 2(l+w)
i.e.
P = 2(l+w)
Here
Given that the length and width of the playground by a scale factor of 2
A scale factor of 2 means we need to multiply both length and width by 2.
i.e
P = 2× 2(l+w)
P' = 2 (2(l+w))
= 2P ∵ P = 2(l+w)
Therefore, we conclude that If Tawnee increases the length and width of the playground by a scale factor of 2, the perimeter of the new playground will be twice the perimeter of the original playground.
