Answer: 50%
Step-by-step explanation:
Answer:
The diameter of the circle O is <em>12 units</em>
Step-by-step explanation:
There is no data in the image provided. To better help you, I'm assuming we have an arbitrary value of
and the triangle B0C is right
<u>Relations in the Circle</u>
The diameter (D) is twice the radius (r) and the radius is the distance measured from the center of the circle to any point of the circumference.
Since the triangle B0C has a right angle, BD is the hypotenuse and 0B=0C=r
Applying Pythagoras's theorem:
Thus we have
The diameter of the circle O is D=2(6) = 12 units
(4x-3y)^2(4x-3y)
=(16x^2-24xy+9y^2)(4x-3y)
=64x^3-96x^2y+36xy^2-48x^2y+72xy^2-27y^3
=64x^2-144x^2y+108xy^2-27y^3
Answer:
19ft
Step-by-step explanation:
Given the height of a ball above the ground after x seconds given by the quadratic function y = -16x2 + 32x + 3, we can find the maximum height reached by the ball since we are not told what to look for.
The velocity of the ball is zero at maximum height and it is expressed as:
V(x) = dy/dx
V(x) = -32x+32
Since v(x) = 0
0 = -32x+32
32x = 32
x = 32/32
x = 1s
Get the height y
Recall that y = -16x² + 32x + 3.
Substitute x = 1
y = -16(1)²+32(1)+3
y = -16+32+3
y = -16+35
y = 19ft
Hence the maximum height reached by the ball is 19ft
Answer:The answer is (x,y)->(+4,y-5)
Step-by-step explanation:
since -4 to 0 is + 4 and 2 to -3 is -5
