I need help with these problems for a test.
1. Find the inverse of the one-to-one function.
f(x)=5x+1/4.
2. The height of the water, H, in feet, at a boat dock t hours after 7 am is given by E=7+4.4cos pi/35 t, where t is time measured in seconds. Find the period.
3. Solve the equation on the interval [0, 2ë).
a) sin^2x - cos^2x = 0
b) sin4x=(sq3)/2
4. Factor completely, or state that the polynomial is prime.
27y^4-147y^2
#3 solve on the interval [0,2pi]
4. 27Y^4-147Y^2=3Y^2(9Y^2-49)=
3Y^2(3Y+7)(3Y-7)
Hint: 9y^2-49 is the difference of 2 squares and is factored into 2 binomials.
F(x)=5X+1/4 Or Y=5X+1/4
To determine the inverse of a function,
replace all Xs with Y and all Ys with X: X=5Y+1/4 Then Multiply both sides of eguation by 4 to eliminate the fraction: 4X=20Y+1. Solve for Y:
Y=(4X-1)/20= Inverse function.
#1) 3y + 12 / 8y third power divided 9y + 36 / 16y third power #2) 8x second power / x second power - 9 x 2 second power + 6x +9 / 16x third power
f(x)=3-4x when f(x)=-5
Sure, I can help you with these problems. Let's break them down one by one.
1. To find the inverse of a one-to-one function, you need to switch the roles of x and y in the equation and solve for y. Let's find the inverse of the function f(x) = 5x + 1/4.
Switching the roles of x and y, we have:
x = 5y + 1/4
Now, we solve for y:
x - 1/4 = 5y
(x - 1/4) / 5 = y
Therefore, the inverse function is:
f^(-1)(x) = (x - 1/4) / 5
2. The equation given is E = 7 + 4.4cos(pi/35 * t), where E represents the height of the water and t represents time in seconds.
To find the period of a trigonometric function, you need to find the smallest positive value of t for which the function repeats itself. In this case, we want to find the smallest positive value of t for which cos(pi/35 * t) repeats itself.
The general formula to find the period of a cosine function is:
Period = 2π / abs(angular frequency)
For the given function, the angular frequency is pi/35. Therefore, the period is:
Period = 2π / (pi/35) = 70
So, the period of the function is 70 seconds.
3. Let's solve the equations on the given interval [0, 2π).
a) To solve sin^2x - cos^2x = 0, we can use the trigonometric identity sin^2x - cos^2x = sin(2x) = 0.
Setting sin(2x) = 0, we have:
2x = 0 or 2x = π
For the interval [0, 2π), the solutions are x = 0 and x = π/2.
b) To solve sin4x = √3/2, we use the inverse sine function to find the values of x.
sin^(-1)(√3/2) = π/3
Since the equation sin4x = √3/2, we have:
4x = π/3, 4x = -π/3 + 2π, or 4x = π/3 + 2π
Simplifying, we find:
x = π/12, x = -π/12 + π/2, or x = π/12 + π/2
For the interval [0, 2π), the solutions are x = π/12 and x = 5π/12.
4. We need to factor the polynomial 27y^4 - 147y^2.
Let's first look for common factors. We notice that both terms have a common factor of 27y^2.
So we can rewrite the polynomial as:
27y^2(y^2 - 7)
Now, let's check if y^2 - 7 can be further factored. In this case, it cannot be factored any further because it is in its simplest form.
Therefore, the completely factored form of the polynomial is:
27y^2(y^2 - 7)
So, the polynomial is not prime.
I hope this helps you with your test preparation! Let me know if you have any further questions.