1. ## simplify radical

i have a few questions that i need to learn how to do.

Simplify
1. a) (2root3 - 3root2)(root5 + 4root3)
b) 2root5 / 3root2
c) root2 - root3 / 2root3 + 3root2

Factoring:
1. a) 5y - 40y^4
b) 54y^6 + 16x^3
c) 2x^2(2x + 1) - x(2x + 1) + 3(2x + 1)
d) 80x^4 - 45x^2y^2

Trigonometric Functions

1. A periodic function f has a period of 15. If f(5) = 13 and f(40 = -5, determine f(110).

2. An object suspended from a spring moves up and down. The distance from max point to low point is 30cm and it takes 4s to complete 5 cycles. Model this equation as a sine function, describing distance from mean position d(t) cm, with respect to time (t s).

Logarithmic Functions
1. Express x as a logarithmic function of y.
y = 2.7(8)^x

it will be so helpful if you could help me, even if its just a few of the questions. thank you!!!

2. Originally Posted by checkmarks

Simplify
1. a) (2root3 - 3root2)(root5 + 4root3)
Just expand the brackets regularly, you do know how to do that right?

$\left( 2 \sqrt {3} - 3 \sqrt {2} \right) \left( \sqrt {5} + 4 \sqrt { } \right) = 2 \sqrt {3} \sqrt {5} + 8 \sqrt {3} \sqrt {3} - 3 \sqrt {2} \sqrt {5} - 12 \sqrt {2} \sqrt {3}$

$= 72 + \left( 2 \sqrt {3} - 3 \sqrt {2} \right) \sqrt {5} - 12 \sqrt {6}$

b) 2root5 / 3root2
$\frac {2 \sqrt {5}}{ 3 \sqrt {2}} = \frac { \sqrt {20}}{ \sqrt {18}} = \frac { \sqrt {10} \sqrt {2}}{ \sqrt {9} \sqrt {2}} = \frac { \sqrt {10}}{3}$

c) root2 - root3 / 2root3 + 3root2
$\frac { \sqrt {2} - \sqrt {3}}{ 2 \sqrt {3} + 3 \sqrt {2}} = \frac { \sqrt {2} - \sqrt {3}}{ 2 \sqrt {3} + 3 \sqrt {2}} \cdot \frac { 2 \sqrt {3} - 3 \sqrt {2}}{ 2 \sqrt {3} - 3 \sqrt {2}}$

$= \frac {2 \sqrt {6} - 6 - 6 + 3 \sqrt {6}}{12 - 18}$

$= \frac {5 \sqrt {6} - 12}{-6}$

$= \frac {12 - 5 \sqrt {6}}{6}$

Logarithmic Functions
1. Express x as a logarithmic function of y.
y = 2.7(8)^x
Recall the definition of a logarithm.

If $\log_a b = c$
then $a^c = b$

So we have:

$y = 2.7(8)^x$

$\Rightarrow \frac {y}{2.7} = 8^x$

$\Rightarrow \log_8 \left( \frac {y}{2.7} \right)= x$

OR

$\log_8 y - \log_8 2.7 = x$

3. Originally Posted by checkmarks
Factoring:
1. a) 5y - 40y^4
$5y - 40y^4 = 5y \left( 1 - 8y^3 \right)$

$= 5y \left( 1^3 - (2y)^3 \right)$

$= 5y (1 - 2y) (1 + 2y + 4y^2)$ .........difference of two cubes

b) 54y^6 + 16x^3
$54y^6 + 16x^3 = 2 \left( 27y^6 + 8x^3 \right)$

$= 2 \left( \left( 3y^2 \right)^3 + (2x)^3 \right)$ .......now this is the sum of two cubes

$= 2 \left( 3y^2 + 2x \right) \left( 9y^4 - 6xy^2 + 4x^2 \right)$

c) 2x^2(2x + 1) - x(2x + 1) + 3(2x + 1)
$2x^2 (2x + 1) - x(2x + 1) + 3 (2x + 1) = (2x + 1) \left( 2x^2 - x + 3 \right)$ ..........factored out the common $(2x + 1)$

this can't be factored anymore, we would need a -3. are you sure the original question wasn't $2x^2 (2x + 1) - x(2x + 1) - 3(2x + 1)$ ?

d) 80x^4 - 45x^2y^2
$80x^4 - 45x^2 y^2 = 5x^2 \left( 16x^2 - 9y^2 \right)$ .....this is the difference of two squares

$= 5x^2 (4x + 3y)(4x - 3y)$

so what did we learn here. when we need to factor something, the first step is to see if there are any common factors we can pull out. then we look if the rest can be factored using things like "the difference of two squares" or "the sum/difference of two cubes" or just regular quadratic factoring

4. thank you soooo mcuh, you have no idea how much that has helped me

i dont know? teh question was written that way.

also, theres one more factoring question im stuck on?
4x^2 + 12xy + 9y^2 - 25

i tried grouping....

5. Originally Posted by checkmarks
thank you soooo mcuh, you have no idea how much that has helped me

i dont know? teh question was written that way. the answer is (2x+1)(2x^2-x+3) but i already thought the question was factored..

also, theres one more factoring question im stuck on?
4x^2 + 12xy + 9y^2 - 25

i tried grouping....
$4x^2 + 12xy + 9y^2 - 25$ ........factor everything but the 25

$\Rightarrow (2x + 3y)(2x + 3y) - 25 = (2x + 3y)^2 - 5^2$ .........this is the difference of two squares

$= [(2x + 3y) + 5][(2x + 3y) - 5]$

$= (2x + 3y + 5)(2x + 3y - 5)$

6. i had my math exam today and it was relatively easy
so thanks for all your help. !

7. Originally Posted by checkmarks
i had my math exam today and it was relatively easy
so thanks for all your help. !
Good going!

-Dan