Answers:
- Part A) There is one pair of parallel sides
- Part B) (-3, -5/2) and (-1/2, 5/2)
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Explanation:
Part A
By definition, a trapezoid has exactly one pair of parallel sides. The other opposite sides aren't parallel. In this case, we'd need to prove that PQ is parallel to RS by seeing if the slopes are the same or not. Parallel lines have equal slopes.
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Part B
The midsegment has both endpoints as the midpoints of the non-parallel sides.
The midpoint of segment PS is found by adding the corresponding coordinates and dividing by 2.
x coord = (x1+x2)/2 = (-4+(-2))/2 = -6/2 = -3
y coord = (y1+y2)/2 = (-1+(-4))/2 = -5/2
The midpoint of segment PS is (-3, -5/2)
Repeat those steps to find the midpoint of QR
x coord = (x1+x2)/2 = (-2+1)/2 = -1/2
y coord = (x1+x2)/2 = (3+2)/2 = 5/2
The midpoint of QR is (-1/2, 5/2)
Join these midpoints up to form the midsegment. The midsegment is parallel to PQ and RS.
What are you referring to with, what is P? Do you mean which crayons were most likely picked?
If so, then blue was most likely picked.
Answer:
2.03:no
2.05:no
2.04:yes
2.01:yes
2.02:yes
Step-by-step explanation:
Answer:
1) Square root
2) Perfect Square
3) Irrational Number
Step-by-step explanation:
You have to estimate the slope of the tangent line to the graph at <em>t</em> = 10 s. To do that, you can use points on the graph very close to <em>t</em> = 10 s, essentially applying the mean value theorem.
The MVT says that for some time <em>t</em> between two fixed instances <em>a</em> and <em>b</em>, one can guarantee that the slope of the secant line through (<em>a</em>, <em>v(a)</em> ) and (<em>b</em>, <em>v(b)</em> ) is equal to the slope of the tangent line through <em>t</em>. In this case, this would be saying that the <em>instantaneous</em> acceleration at <em>t</em> = 10 s is approximately equal to the <em>average</em> acceleration over some interval surrounding <em>t</em> = 10 s. The smaller the interval, the better the approximation.
For instance, the plot suggests that the velocity at <em>t</em> = 9 s is nearly 45 m/s, while the velocity at <em>t</em> = 11 s is nearly 47 m/s. Then the average acceleration over this interval is
(47 m/s - 45 m/s) / (11 s - 9 s) = (2 m/s) / (2 s) = 1 m/s²
