The distance from Humood's house to the school is 500m
<h3>How to determine the distance from Humood's house to the school?</h3>
The given parameters are:
Humood's house is (-5,7)
The school is (3,1)
The distance between both points is calculated using
Substitute the known values in the above equation
Evaluate
d = 10
Each unit in the graph is 50m.
So, we have
Distance =10 * 50m
Evaluate the product
Distance = 500m
Hence, the distance from Humood's house to the school is 500m
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Since they’re vertical we know that they’re congruent. We write the equation as 160=10x+50. Subtract 50 to both sides then we get 110=10x so now we divide 10 to both sides and get 11=x.
So your answer is A
Hope I helped :)
Answer:
the answer is x=-2 as answer
Answer:
The answer is 29
Step-by-step explanation:
11+10+5+3=29
RS => y - 5 = (8 - 5)/(1 - (-1)) (x - (-1))
y - 5 = 3/2 (x + 1) => slope = 3/2
ST => y - 8 = (-2 - 8)/(7 - 1) (x - 1)
y - 8 = -10/6 (x - 1) = -5/3 (x - 1) => slope = -5/3
TU => y - (-2) = (0 - (-2))/(2 - 7) (x - 7)
y + 2 = 2/5(x - 7) => slope = 2/5
UR => y = 5/(-1 - 2) (x - 2)
y = -5/3 (x - 2) => slope = -5/3
The median is the line joining the midpoints of the non-parallel sides.
Midpoint of RS = ((-1 + 1)/2, (5 + 8)/2) = (0, 13/2)
Midpoint of TU = ((7 + 2)/2, -2/2) = (9/2, -1)
Equation of the line joining (0, 13/2) and (9/2, -1) is given by y - 13/2 = (-1 - 13/2)/(9/2) x
y - 13/2 = (-15/2)/(9/2) x
y - 13/2 = -15/9x
18y - 117 = -30x
30x + 18y = 117
