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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Answer:
yes
Step-by-step explanation:
it's saying pretty much take 15 out of 38
Check the picture below.
notice that the triangle ADH, since the segment AL is an angle bisector, meaning it cuts the angle A in two equal halves, then the triangle ADH is only using half of A.
Answer:
∠EFG = 48°
Step-by-step explanation:
As FH bisects ∠EFG , ∠EFH = ∠HFG .
We know that ∠EFH = (-5x + 89)° . So ∠HFG = ∠EFH = (-5x + 89)°
Also, ∠HFG + ∠EFH = ∠EFG
=> 2(-5x + 89)° = (61 - x)°
=> -10x + 178 = 61 - x
=> 10x - x = 178 - 61
=> 9x = 117
=> x = 117 / 9 = 13
Putting the value of 'x' in ∠EFG gives :-
(61 - x)° = (61 - 13)° = 48°
<span>C. (4,3)
Draw and plot on a graph and (4,3) could be the other end point, which
x < 5
y > 2</span>
