$57.60 is the correct answer
Answer:
a
Step-by-step explanation:
sec (thita) = squrt (5)
squaring on both sides:
sec^2 (thita) = 5 - equation 1
1 + tan^2 (thita) = 5
tan^2 (thita) = 4
tan (thita) = 2.
= tan (thita) - squrt(5)sin(thita)
= 2 - squrt(5) x 2/ squrt(5)
= 0
from eqn - 1
sec(thita) = squrt(5)
cos(thita) = 1/ squrt(5)
sin(thita) = squrt ( 1- 1/(squrt (5))^2)
sin(thita) = 2/ squrt(5) .
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:
11.25cm^2
Step-by-step explanation:
given the area of the rectangle as 40cm^2, we can get the width of the rectangle.
Area =Length x width
40cm^2 = 8cm x width
we can divide both sides 8 to be left with the width
40/8 = 5cm
so the width is 5cm which is equal to the height of the triangle.
Area of BDC = 1/2 x base x height
BDC = 1/2 x 4.5 x 5
=11.25cm^2
Answer:
I don't know how to work it out because I have not answered this question
