Since we are dealing with volume, when one object is 4 times bigger than another, the difference in volume is 4^3 = 64.
6 cm sphere = 4/3 *PI * 6^3
6 cm sphere = 4/3 *PI * 216
6 cm sphere =
<span>
<span>
<span>
904.78 cc
24 cm sphere = </span></span></span>4/3 *PI * 24^3
24 cm sphere = 4/3 *PI * 13,824
24 cm sphere =
<span>
<span>
<span>
57,905.84
</span>
</span>
</span>
cc
Comparing the volumes: <span>57,905.84 / 904.78 = </span><span><span><span>64
</span>
</span>
</span>
<span>
</span>
(5,2)(1,0)
slope = (0 - 2) / (1 - 5) = -2 / -4 = 1/2 <===
Answer:
$9,393.78
Step-by-step explanation:
Using the equation:
A = P(1+r)^t
Where,
A = final amount
P = initial amount = $6,600
r = rate of increase = 4% = 0.04
t = time in years = 9 years (2012-2021)
A = 6,600(1 + 0.04)^9
= 6,600(1.04)^9
= 6,600(1.4233)
= 9,393.78
A = $9,393.78
Answer:
45 minutes
Step-by-step explanation:
At 30 mph for 1/4 hour, Peter has a 7.5 mile head start. After he leaves, Mitchell closes that gap at the rate of 40-30 = 10 miles per hour. It will take him ...
t = d/s
t = (7.5 mi)/(10 mi/h) = 0.75 h
to catch Peter.
Mitchell will catch Peter in 45 minutes.
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<em>Alternate Solution</em>
Another way to look at it is that Mitchell's 10 mph advantage is 1/3 of Peter's speed, so it will take 1/(1/3) = 3 times the period of Peter's head start:
3 × 15 minutes = 45 minutes . . . for Mitchell to catch Peter
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You can write equations involving time and distance and see where the distances traveled become the same. You need to be careful choosing the time reference, since you're concerned with Mitchell's travel time. I personally prefer to work "head start" problems by considering the differences in time and speed, as above. This is where you end up using the equations approach, anyway.
Greater than because the line is dashed and the shaded region is in the positive or less negative numbers area. Greater than or equal would be the same but with a solid line.
