# Math Help - Need help with funtions and quadratics

1. ## Need help with funtions and quadratics

My exam is tomorrow and i really need some help with the following questions
If possible plz explain all the steps, cuz there's no point in learning without knowing how to do

I know there are alot of question, but i really need help cuz my exam is in like 20 hours, do as many as possible thanks (All of these questions are from Grade 11 Nelson mathematics Book)

1) Tiffany is paid $8.05/h plus 5% of hr sales over$1000, for a 40-h work week. For example, Suppose Tiffany sold $1800 worth of merchandise. Then she would earn$ 8.05(40)+0.005($800) =$362

1a) Graph the relation between total pay for 40h work week and her sales for the week

1b) Write the relation in function notation

1c) Write the inverse in function notation

2) It costs a bus company $225 to run a minibus on a ski trip, plus$30 per passenger. The bus has seating for 22 passengers, and the company charges $60 per fare if the bus is full. For each empty seat, the company has to increase price by$ 5. How many empty seats should the bus run to maximize the profit from this trip.

3) Given that f(X)= -3x +8, determine (solved by Dan)
a) the value to x where f(X) = 14
b) the value(s) of x where f(x) = f-1 (X)

Determine the equation of g-1 (Solved by Dan)
a) G(X) = 2x^2 +12x -14
b) G(X) = (root of) 4x + 8

4) A ferris wheel has a radius of 10m and completes one full revolution in 36s. the riders board the ride from a platform
1m above the ground at the bottom of the wheel (Solved by Dan)
a) Determine and equation that models the position of a rider above the ground, at time, t, in seconds.

b) Determine rider's height above the ground 10 seconds after the ride starts

C) When will the rider be at max height? what is this height

6) 9= 4 tan 2x 0 degree less than or equal to x less than or equal to 360 degree (Solved by Dan)

Thanks

2. Originally Posted by supersaiyan
3) Given that f(X)= -3x +8, determine
a) the value to x where f(X) = 14
b) the value(s) of x where f(x) = f-1 (X)
a)
[tex]14 = -3x + 8[tex]

$3x = 8 - 14 = -6$

$x = -2$

b) I presume that "f-1 (X)" is supposed to be the inverse function?

$x = f(y)$ gives the inverse function.

$x = -3y + 8$

$3y = -x + 8$

$y = f^{-1}(x) = \frac{-x + 8}{3}$

So when is $\frac{-x + 8}{3} = -3x + 8$?

$-x + 8 = -9x + 24$

$8x = 24 - 8 = 3$

$x = \frac{3}{8}$

-Dan

3. Originally Posted by supersaiyan
Determine the equation of g-1
a) G(X) = 2x^2 +12x -14
b) G(X) = (root of) 4x + 8
a) $g(x) = 2x^2 + 12x - 14$

Let's "clean this up" a little:
$g(x) = (2x^2 + 12x) - 14$

$g(x) = 2(x^2 + 6x) - 14$

$g(x) = 2(x^2 + 6x + 9 - 9) - 14$

$g(x) = 2(x^2 + 6x + 9) - 18 - 14$

$g(x) = 2(x + 3)^2 - 32$

Now, to find the inverse function, let $y = g(x) = 2(x + 3)^2 - 32$. The inverse function will be given by $x = g(y) = 2(y + 3)^2 - 32$.

$x = 2(y + 3)^2 - 32$

$x + 32 = 2(y + 3)^2$

$(y + 3)^2 = \frac{x + 32}{2}$

$y + 3 = \sqrt{\frac{x + 32}{2}}$

$y = -3 + \sqrt{\frac{x + 32}{2}}$

b)
$g(x) = \sqrt{4x + 8}$

Again, the inverse function is given by
$x = \sqrt{4y + 8}$

$x^2 = 4y + 8$

$4y = x^2 - 8$

$y = \frac{x^2 - 8}{4}$

-Dan

4. thanks alot dan for your solutions

5. Originally Posted by supersaiyan
4) A ferris wheel has a radius of 10m and completes one full revolution in 36s. the riders board the ride from a platform 1m above the ground at the bottom of the wheel
a) Determine and equation that models the position of a rider above the ground, at time, t, in seconds.

b) Determine rider's height above the ground 10 seconds after the ride starts

C) When will the rider be at max height? what is this height
Don't they teach the spelling of "pi" in school anymore?? If I had done that in High School they'd've taken points off.

First draw a sketch. The center of the ferris wheel is 1 m + 10 m = 11 m off the ground. The rider gets on at t = 0 s where I will call $\theta = 0$ rad. And I'm assuming the ferris wheel is rotating counterclockwise. I'm setting up a coordinate system at ground, right at the point where the rider gets on the ferris wheel. +x is to the right, and +y is directly upward.

The wheel completes one turn in 36 s, so the rotational frequency is $\frac{2 \pi~rad}{36~s} = \frac{\pi}{18}~rad/s$.

a) I am going to give the equation of motion parametrically in terms of time.

We have:
$x = 10 \cdot cos(\omega t + \pi/2) = 10 \cdot cos \left ( \frac{\pi t}{18} + \frac{\pi}{2} \right )$

$y = 11 - 10 \cdot sin(\omega t + \pi/2) = 11 - 10 \cdot sin \left ( \frac{\pi t}{18} + \frac{\pi}{2} \right )$
(The $\pi/2$ is a term to offset the fact we are using the motion equations starting from the "straight down" position.)

b) t = 10 s, so
$y = 11 - 10 \cdot sin \left ( \frac{\pi (10)}{18} + \frac{\pi}{2} \right ) \approx 12.7365~m$

c) Max height will be when the position of the rider is directly above the starting point, 1 m + 10 m + 10 m = 21 m in the air. The quick way to find this is to find when
$\frac{\pi t}{18} = \pi$, or at t = 18 s.

The slower, but more general way is to solve:
$21 = 11 - 10 \cdot sin \left ( \frac{\pi t}{18} + \frac{\pi}{2} \right )$

$10 = -10 \cdot sin \left ( \frac{\pi t}{18} + \frac{\pi}{2} \right )$

$-1 = sin \left ( \frac{\pi t}{18} + \frac{\pi}{2} \right )$

Thus
$\frac{\pi t}{18} + \frac{\pi}{2} = \frac{3 \pi}{2}$ <-- The first time this happens, anyway.

$\frac{\pi t}{18} = \pi$

which again gives t = 18 s.

-Dan

6. Originally Posted by supersaiyan
5) 0.5 sin (theta), 0 less than or equal to theta less that or equal to 2 pie

6) 9= 4 tan 2x 0 degree less than or equal to x less than or equal to 360 degree
What is 5??

6) Solve $9 = 4 \cdot tan(2x)$ for $0 \leq x < 2 \pi$.

$tan(2x) = \frac{9}{4}$

$2x = tan^{-1} \left ( \frac{9}{4} \right )$

$x = \frac{1}{2} \cdot tan^{-1} \left ( \frac{9}{4} \right ) \approx 0.576286$

We know that tan(2x) is periodic every $\pi/2$, so we may add integral multiples of $\pi/2$ to x:
$x = \frac{1}{2} \cdot tan^{-1} \left ( \frac{9}{4} \right ), \frac{1}{2} \cdot tan^{-1} \left ( \frac{9}{4} \right ) + \frac{\pi}{2}$ $, \frac{1}{2} \cdot tan^{-1} \left ( \frac{9}{4} \right ) + \pi, \frac{1}{2} \cdot tan^{-1} \left ( \frac{9}{4} \right ) + \frac{3 \pi}{2}$ rad

or
$x \approx 0.576286, 2.14708, 3.71788, 5.28867$ rad.

-Dan

7. Thanks again dan

Now i only need solutions for number 1 and 2

i figured out 5