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:
3y+1
Step-by-step explanation:
4y+3-y-2
3y+3-2
3y+1
1 + 4 = 5 (5 parts)
<span>£250/5 = 50 </span>
<span>1 x £50 = 50 </span>
<span>4 x £50 = 200 </span>
<span>so the answer is 50:200 </span>
Answer:
528in
Step-by-step explanation:
SA formula: 2Lw + 2Lh + 2hw
for this box: 2(4×10)+2(10×16)+2(4×16)
2(40)+2(160)+2(64)
80+320+128
80+448
<u>528in</u>