Answer:
After 4 seconds, the ball is at 229 ft
Step-by-step explanation:
Here, we want to get the height of the ball at 4 seconds
What we have to do here is to substitute the value of t for 4 seconds in the equation
We have;
h = -16(4^2) + 121(4) + 1
h = -256 + 484 + 1
h = 229 ft
To solve the problem shown above it is important to know that a binomial is a polynomial that has two terms. When the highest exponent of the binomial is 2, it is a binomial with degree 2, this is also called or know as "quadratic binomial". Therefore, keeping this on mind, you must identify which of the polynomial above has degree 2.
As you can see, the answer is: <span>−6x−1 7x²</span>
We know that the height of the building is 45m, and the
distance between the building and victor is 20m. Since the problem does not
state the height of Victor, we can assume that horizontal line of sight of
Victor coincides with the base of the building. This gives us a right triangle
with angle x and sides 45m and 20m, as you can see in the diagram.
Now, to find the value of the angle x, we will need a
trigonometric function that relates the opposite side of our angle x with the
adjacent side of it; that trigonometric function is tangent. Remember that 
We know for our diagram that the opposite side of Victor's angle of inclination, x, is the height of the building (45m), and the adjacent side of it is the distance between Victor and the building (20m). Now we can replace the values in our tangent equation to get:
But we need to find the value of x not the value of tangent, so we are going to use the inverse function of tangent, arctangent (arctan)
to solve the equation for x:
We can conclude that Victor's angle of inclination from he stands to the top of the building is 66°.
Answer:
Its C the correct one
Step-by-step explanation:
Solve cos(4x)-cos(2x)=0 ∀ 0<=x<=2pi ..............(0)
Normal solution:
1. use the double angle formula to decompose, and recall cos^2(x)+sin^2(x)=1
cos(4x)=cos^2(2x)-sin^2(2x)=2cos^2(2x)-1 .................(1)
2. substitute (1) in (0)
2cos^2(2x)-1-cos(2x)=0
3. substitute u=cos(2x)
2u^2-u-1=0
4. Solve for x
factor
(u-1)(u+1/2)=0
=> u=1 or u=-1/2
However, since cos(x) is an even function, so solutions to
{cos(2x)=1, cos(-2x)=1, cos(2x)=-1/2 and cos(-2x)} ...........(2)
are all solutions.
5. The cosine function is symmetrical about pi, therefore
cos(-2x)=cos(2*pi-2x),
solution (2) above becomes
{cos(2x)=1, cos(2pi-2x)=1, cos(2x)=-1/2, cos(2pi-2x)=-1/2}
6. Solve each case
cos(2x)=1 => x=0
cos(2pi-2x)=1 => cos(2pi-0)=1 => x=pi
cos(2x)=-1/2 => 2x=2pi/3 or 2x=4pi/3 => x=pi/3 or 2pi/3
cos(2pi-2x)=-1/2 => 2pi-2x=2pi/3 or 2pi-2x=4pi/3 => x=2pi/3 or x=4pi/3
Summing up,
x={0,pi/3, 2pi/3, pi, 4pi/3}
