If you do 22÷9 and see the hole number: 2. Then do 2 x 9 and you get 18. Now do 22-18 and you get 4. So do 2 4/9. Always have the denominator stay the same.
Answer:
2 hours, 150 miles
Step-by-step explanation:
The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.
<h3>Separation distance</h3>
Jason got a head start of 20 miles, so that is the initial separation between the two drivers.
<h3>Closure speed</h3>
Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.
<h3>Closure time</h3>
The relevant relation is ...
time = distance/speed
Then the time it takes to reduce the separation distance to zero is ...
closure time = separation distance / closure speed = 20 mi / (10 mi/h)
closure time = 2 h
Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.
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<em>Additional comment</em>
The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.
Step-by-step explanation:
let's look at the full numbers under the square roots when bringing the external factors back in :
sqrt(9×9×2) - sqrt(3×3×7) + sqrt(8) - sqrt(28)
and let's present these numbers as the product of their basic prime factors
sqrt(3×3×3×3×2) - sqrt(3×3×7) + sqrt(2×2×2) - sqrt(2×2×7)
now we see that we have 2 pairs of square roots : 1 pair ends with a factor of 2, and one pair with a factor of 7.
let's combine these
sqrt (3×3×3×3×2) + sqrt(2×2×2) - sqrt(3×3×7) - sqrt (2×2×7)
and now we move the factors of 2 and 7 back out in front (of course, we need to apply the square root on these factors) :
9×sqrt(2) + 2×sqrt(2) - 3×sqrt(7) - 2×sqrt(7) =
= (9+2)×sqrt(2) - (3+2)×sqrt(7) = 11×sqrt(2) - 5×sqrt(7)
and that is the first answer option.
I believe the answer to your question is b.