Answer:
Step-by-step explanation:
Given the equation
Sin(5x) = ½
5x = arcSin(½)
5x = 30°
Then,
The general formula for sin is
5θ = n180 + (-1)ⁿθ
Divide through by 5
θ = n•36 + (-1)ⁿ30/5
θ = 36n + (-1)ⁿ6
The range of the solution is
0<θ<2π I.e 0<θ<360
First solution
When n = 0
θ = 36n + (-1)ⁿθ
θ = 0×36 + (-1)^0×6
θ = 6°
When n = 1
θ = 36n + (-1)ⁿ6
θ = 36-6
θ = 30°
When n = 2
θ = 36n + (-1)ⁿ6
θ = 36×2 + 6
θ = 78°
When n =3
θ = 36n + (-1)ⁿ6
θ = 36×3 - 6
θ = 102°
When n=4
θ = 36n + (-1)ⁿ6
θ = 36×4 + 6
θ = 150
When n =5
θ = 36n + (-1)ⁿ6
θ = 36×5 - 6
θ = 174°
When n = 6
θ = 36n+ (-1)ⁿ6
θ = 36×6 + 6
θ = 222°
When n = 7
θ = 36n + (-1)ⁿ6
θ = 36×7 - 6
θ = 246°
When n =8
θ = 36n + (-1)ⁿ6
θ = 36×8 + 6
θ = 294°
When n =9
θ = 36n + (-1)ⁿ6
θ = 36×9 - 6
θ = 318°
When n =10
θ = 36n + (-1)ⁿ6
θ = 36×10 + 6
θ = 366°
When n = 10 is out of range of θ
Then, the solution is from n =0 to n=9
So the equation have 10 solutions in the range 0<θ<2π
From the given function modeling the height of the ball:
f(x)=-0.2x^2+1.4x+7
A] The maximum height of the ball will be given by:
At max height f'(x)=0
from f(x),
f'(x)=-0.4x+1.4
solving for x we get:
-0.4x=-1.4
x=3.5ft
thus the maximum height would be:
f(3.5)=-0.2(3.5)^2+1.4(3.5)+7
f(3.5)=9.45 ft
b]
How far from where the ball was thrown did this occur:
from (a), we see that at maximum height f'(x)=0
f'(x)=-0.4x+1.4
solving for x we get:
-0.4x=-1.4
x=3.5ft
This implies that it occurred 3.5 ft from where the ball was thrown.
c] How far does the ball travel horizontally?
f(x)=-0.2x^2+1.4x+7
evaluationg the expression when f(x)=0 we get:
0=-0.2x^2+1.4x+7
Using quadratic equation formula:
x=-3.37386 or x=10.3739
We leave out the negative and take the positive answer. Hence the answer 10.3739 ft horizontally.
Since the area of a square is equal to the square of one of its side's length, then the area should be equivalent to
.
---> equation (1)
By using pythagoras rule which states that the
---> equation (2)
where the opposite side's length is 8 and the hypotenuse side's length is 10
by substituting by the values in equation (2) therefore,
substitute this value in equation (1) then
where A is the area of the square whose side is x
