1. You have that:
- The homeowner<span> want the length of the swimming pool to be 4 feet longer than its width.</span>
- He wants to surround it with a concrete walkway 3 feet wide.
- He can only afford 300 square feet of concrete for the walkway.
2. Therefore, the tota area is:
At=L2xW2
L2 is the lenght of the walkway (L2=L1+3+3⇒L2=(W1+4+6)⇒L2=W1+10).
W2 is the width of the walkway (W1+3+3⇒W2=W1+6)
3. The area of the walkway is:
A2=At-A1
A2=300 ft²
4. Therefore, you have that the width of the swimming pool is:
A2=(W1+10)(W1+6)-(W1+4)(W1)
300=(W1²+6W1+10W1+60)-(W1²+4W1)
W1²+16W1+60-W1²-4W1-300=0
12W1-240=0
W1=240/12
W1=20 ft
5. And the length is:
L1=W1+4
L1=20+4
L1=24 ft
Answer: 1/5 or 8 houses. Company X provides electricity to 8 houses, Company Y provides to 24 houses, so that would mean Company Z provides 8
<1 = 120/2 = 60
answer
60 <span>degrees</span>
Answer:
2x+10
hope this helps
have a good day :)
Step-by-step explanation:
Check the picture below.

since the distance AB cannot be a negative value, thus is not -3-√(130).