1. If y<span>ou have jogged 5 miles from the park at a rate of r miles per hour, then the time you need to do this is
![\frac{5}{r}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7Br%7D%20)
hours (
![S=v\cdot t](https://tex.z-dn.net/?f=S%3Dv%5Ccdot%20t)
, where S is distance, v-speed and t-time).
</span>
2. W<span>hen you run back to the park, your average speed increases by 1 mph and become r+1 mph.
</span>
3. I<span>t takes
![\frac{5}{r+1}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7Br%2B1%7D%20)
hours to jog 5 miles back to the park</span>
4. T<span>he total jogging time T is
![\frac{5}{r} + \frac{5}{r+1}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7Br%7D%20%2B%20%5Cfrac%7B5%7D%7Br%2B1%7D%20)
hours.</span>
<span />
5. If you jogged away from the park at an average speed of 4 miles per hour, then r+1=4 and r=3. The expression
![\frac{5}{r} + \frac{5}{r+1}](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7Br%7D%20%2B%20%5Cfrac%7B5%7D%7Br%2B1%7D%20)
will take look
<span>
![\frac{5}{3} + \frac{5}{4}=\frac{5\cdot4+5\cdot 3}{12} =\frac{35}{12}=2.9](https://tex.z-dn.net/?f=%20%5Cfrac%7B5%7D%7B3%7D%20%2B%20%5Cfrac%7B5%7D%7B4%7D%3D%5Cfrac%7B5%5Ccdot4%2B5%5Ccdot%203%7D%7B12%7D%20%3D%5Cfrac%7B35%7D%7B12%7D%3D2.9%20)
hours.</span>
Answer:
Write 6.384 correct to 1 decimal place ?=6
Answer:
average rate of change = 30
Step-by-step explanation:
the average rate of change of f
(
x)
over an
interval between 2 points is the slope of the secant
line connecting the 2 points
It is calculated as
f
(
b
)
−
f
(
a
)
b
−
a
where a, b is the closed interval
[
a
,
b
]
here
[
a
,
b
]
=
[
1
,
5
]
f
(
b
)
=
f
(
5
)
=
5
3
−
5
=
120
f
(
a
)
=
f
(
1
)
=
1
−
1
=
0
⇒
120
−
0
5
−
1
=
120
4
=
30
Answer:272
Step-by-step explanation:
68•4=272
the decimal is the center to the left is bigger to the right is smaller.
thousand hundreds tens ones (decimal) tenths hundreths thousandths