9514 1404 393
Answer:
pool radius is 8+4√5 ≈ 16.94 meters
Step-by-step explanation:
Let r represent the radius of the pool. Then r+2 is the radius to the outside of the walkway. The ratio of areas is given as 4/5, so we have ...
πr²/(π(r+2)²) = 4/5
5r² = 4(r² +4r +4) . . . . cross-multiply
r² -16r -16 = 0 . . . . . . . subtract the right-side expression from both sides
__
r² -16r +64 = 80 . . . . . add 80 to complete the square
(r -8)² = 80
r -8 = √80 = 4√5 . . . . square root; use only the positive root
r = 8 +4√5 ≈ 16.94 . . . meters
The radius of the pool is 8+4√5 ≈ 16.94 meters.
_____
The area of a circle of radius r is given by the formula ...
A = πr²
Plug the numbers into the equation
5(5) - 2 = ?
25 -2 = 23
23
Answer:
Below in bold.
Step-by-step explanation:
A. −16x2 + 24x + 16 = 0
-8(2x2 - 3x - 2) = 0
-8(2x + 1 )(x - 2) = 0
x = -0.5, 2.
So the x-intercepts are (-0/5, 0) and (2, 0).
B. As the leading coefficients is negative (-16) the vertex of the graph will be a Maximum.
To find its coordinates we convert the function to vertex form:
f(x) = −16x2 + 24x + 16
= -16(x^2 - 1.5x) + 16
Completing the square on contents of the parentheses:
= -16 [(x - 0.75)^2 - 0.75^2] + 16
= -16(x - 0.75)^2 - 16 * -0.75^2 + 16
= -16(x - 0.75)^2 + 9 + 16
= -16(x - 0.75)^2 + 25.
So the coordinates of the vertex are (0.75, 25)
Answer:
Step-by-step explanation:
Take a triangle ABC, in which AB=AC.
Construct AP bisector of angle A meeting BC at P.
In ∆ABP and ∆ACP
AP=AP[common]
AB=AC[given]
angle BAP=angle CAP[by construction]
Therefore, ∆ABP congurent ∆ACP[S.A.S]
This implies, angle ABP=angleACP[C.P.C.T]
Sub 5 for x
x2+4
5*2+4
Gemdas or Pemdas
10+4=14 so the value is 14.
