<u>Answer-</u>
<em>Strip with </em><em>2.13 ft </em><em>width can be used. </em>
<u>Solution-</u>
A club swimming pool is 21 ft wide and 41 ft long.
Hence, the perimeter of the swimming pool is
The club members want a uniform width border around the pool.
They have enough material for 264 ft²
As they will cover the perimeter of the rectangular pool, so the length of the total used strip is equal to the perimeter of the swimming pool.
Let the width of the strip is x ft
So, the total area of the used strip is,
Putting the values,
Therefore, strip with 2.13 ft width can be used.
Slope form: y=mx + b
slope is 2/5, fill into equation
Y = 2/5x + b
Plug in point to find b
1 = 2/5(0) + b
1 = 0 + b, b = 1
The answer is A. Y = 2/5x + 1
Answer:
506.7 cm²
Step-by-step explanation:
The amount of cardboard needed would be the surface area of the cone,
S= pi(r)(l+r) where r is the radius and l is the slant height.
You can find l using the Pythagorean theorem. l²=r²+h² which is l²=6²+20² and you get l = √436.
Plug r and l into the equation: S = 6pi(√436 + 6) = 506.7 cm²
Let me know if you have any questions about my steps!
4 Hope this helped ;):):)
Answer:
rn or later?
Step-by-step explanation:
