Average speed for the entire trip, both ways, is
(Total distance) divided by (total time) .
We don't know the distance from his house to the gift store,
and we don't know how long it took him to get back.
We'll need to calculate these.
-- On the trip TO the store, it took him 50 minutes, at 6 mph.
-- 50 minutes is 5/6 of an hour.
-- Traveling at 6 mph for 5/6 of an hour, he covered 5 miles.
-- The gift store is 5 miles from his house.
-- The total trip both ways was 10 miles.
-- On the way BACK home from the store, he moved at 12 mph.
-- Going 5 miles at 12 mph, it takes (5/12 hour) = 25 minutes.
Now we have everything we need.
Distance:
Going: 5 miles
Returning: 5 miles
Total 10 miles
Time:
Going: 50 minutes
Returning: 25 minutes
Total: 75 minutes = 1.25 hours
Average speed for the whole trip =
(total distance) / (total time)
= (10 miles) / (1.25 hours)
= (10 / 1.25) miles/hours
= 8 miles per hour
Answer:
395.2 cm²
Step-by-step explanation:
2x12.7x6.72=170.688
2x6.72x5.78=77.6832
2x5.78x12.7=146.812
170.688+77.6832+146.812=395.1832
Answer:
(d) 112 m²
Step-by-step explanation:
The area of a parallelogram is given by the formla ...
A = bh . . . . . . where b is the base length and h is the height
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Here, the base is divided into parts, but their total length is 5m+9m = 14m. The height is given as 8m, so the area is ...
A = (14 m)(8 m) = 112 m²
The area of the parallelogram is 112 square meters.
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<em>Alternate solution</em>
You can also figure this as the sum of the areas of the two triangles and that of the center rectangle.
A = 2(1/2bh) +bh
A = 2(1/2(5 m)(8 m)) +(9 m)(8 m) = 40 m² +72 m² = 112 m²
You can find the segment congruent to AC by finding another segment with the same length. So first, you need to find the length of AC.
C - A = AC
0 - (-6) = AC Cancel out the double negative
0 + 6 = AC
6 = AC
Now, find another segment that also has a length of 6.
D - B = BD
2 - (-2) = BD Cancel out the double negative
2 + 2 = BD
4 = BD
4 ≠ 6
E - B = BE
4 - (-2) = BE Cancel out the double negative
4 + 2 = BE
6 = BE
6 = 6
So, the segment congruent to AC is B. BE .
