Let x = the width of the border
2x+23) by (2x + 33) are the overall dimensions of the pool and the border. Use the area to find x!
Overall area - pool area = 660 sq/ft (aggregate area)
(2x+23)*(2x+33) - (23*33) = 660
Solve.
4x^2+112x = 660
4x^2+112x−660=0 → subtract 660 on both sides
4(x−5)(x+33)=0 → factor
x−5=0 or x+33=0 → set both factors equal to 0
x = 5 or x = -33
In this case, we would use the positive solution, so the strip can be 5 ft. wide!
I hope this helps :)
If bob gets Candy then George does homework
Answer:
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Step-by-step explanation:
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The solution to the proportion X is 30.
Answer:
1) x = 40° & y = 50°
2) x = 100° & y = 80°
Step-by-step explanation:
1)
ABCD is a cyclic quadrilateral (∵ all the vertices of the quadrilateral are touching on the circle). As it is a cyclic quadrilateral , sum of it's opposite angles will be 180°.
⇒ ∠ADC + ∠ABC = 180°
⇒ 130° + y = 180°
⇒ y = 180 - 130 = 50°
In ΔABC ,
∠ACB = 90° (∵ AB is the diameter of the circle and a diameter subtends an angle of 90° on any point on circle.)
Using angle sum property of triangle ,
∠ABC + ∠ACB + ∠CAB = 180°
⇒ y + 90° + x = 180°
⇒ x + 50° + 90° = 180°
⇒ x + 140° = 180°
⇒ x = 180 - 140 = 40°
2)
ΔABC is an isosceles triangle (∵AB = AC). As it is an isosceles triangle , it's base angles will be equal. So , ∠ABC = ∠ACB = 50°
Using angle sum property of triangle ,
∠ABC + ∠ACB + ∠BAC = 180°
⇒ 50° + 50° + y = 180°
⇒ y + 100° = 180°
⇒ y = 180 - 100 = 80°
ABEC is a cyclic quadrilateral (∵ all the vertices of the quadrilateral are on the circle.). As it is a cyclic quadrilateral , sum of it's opposite angles will be 180°.
⇒ ∠BAC + ∠BEC = 180°
⇒ y + x = 180°
⇒ x + 80° = 180°
⇒ x = 180 - 80 = 100°