There are a lot of problems here, if you need help on each one specifically, perhaps some tutoring is in order (I am available for online sessions).
Here are some hints to try them on your own:
16. To simplify \sqrt(12) we factor out any perfect squares from 12...in this case 12=4*3, and 4 is a perfect square...so \sqrt(12) simplifies to \sqrt(4)*\sqrt(3)=2\sqrt(3).
If we have 4x-x we can factor out x and write this as (4-1)x=5x...so if x=\sqrt(10) we know what to do in b.
17. To rationalize a denominator we multiply by 1=[\sqrt(x)] / [\sqrt(x)], where the denominator we are trying to rationalize is \sqrt(x)
18. When we have \sqrt(x)=\sqrt(y) we note that squaring both sides of an equality, preserves the equality...thus
[\sqrt(x)]^2 = [\sqrt(y)]^2, but that just says that x=y...for a. first add 8 to both sides then square...for b. square first then subtract 2x from both sides.
19. and 20. Pythagoras tells us that a^2+b^2=c^2 in any right triangle with hypotenuse: c and legs: a and b
Substitute the values given and solve these by applying a square root to both sides (after moving things around appropriately) where necessary.
21. use calculator
22. To do long division of polynomials we write:
(d+3) | 3d^2+2d-29
and notice first that the largest term of our quotient must be 3d so that d*3d = 3d^2 the largest term (by degree) of the numerator. we note that 3*3d = 9d and so (d+3)*3d = 3d^2+9d...we then subtract this from the numerator...
(3d^2+2d-29) - (3d^2+9d) = -7d-29...
we next observe that we need to multiply (d+3) by -7 to get the -7d term, this gives us -7d-21, and we subtract this off from what is left... (-7d-29) - (-7d-21) = -8...this is our remainder since (d+3) doesn't go into 8 a whole number of times...
Thus (3d^2+2d-29)/(d+3)=3d-7+8/(d+3)...
There is short hand notation for this in long division (and something called synthetic divisions) which can organize this process as follows:
3d - 7
(d+3) | 3d^2+2d-29
3d^2+9d
-7d-29
-7d-21
-8 = remainder
Synthetic division involves eliminating the variable and working with just the coefficients.
23. a. recognize that the denominator has a common factor of 2 in both terms...2c-8=2(c-4)=-2(4-c) and then cancel the numerator out of the denominator...you should get -1/2. b. we should try and factor these two quadratic functions and look for any like terms in the numerator and denominator that can cancel:
In general when we have a quadratic function Ax^2+Bx+C, if we know the roots (say by using the quadratic formula) are a_1 and a_2, then Ax^2+Bx+C = A(x-a_1)(x-a_2).
24. We can multiply the whole equation by the common denominator of the LHS...3(m-4)...yielding:
9+1=18(m-4)/m...next we can multiply the whole equation by the denominator of the RHS...m
This will produce a linear equation in m that can be solved by grouping all terms with m on one side and all constant terms on the other side and dividing by the coefficient of m.
25. y=5/(x-4)+1
First recognize that x=4 is not a valid input into this function y=f(x)=5/(x-4) + 1
What this means is that x can get as close to 4 as we want, but can never actually equal 4...when we divide 5 by numbers close to 0 (when x is near 4, x-4 is near 0), this blows up to +/- INFINITY, and adding one doesn't really change this fact much...so we get a VERTICAL ASYMPTOTE at x=4 in the graph...
When we consider x getting very LARGE (in magnitude...going to +INFINITY or -INFINITY), we notice that 5/(x-4) goes to 0 in both cases...thus the function will get close to y=1 when x is far to the right or left in the graph...this means we have a HORIZONTAL ASYMPTOTE at y=1 that describes the behavior of the graph far to the left and right along the x-axis.
I hope this helps...if you need more clarification, let's setup a tutoring session!
Casey W.
06/05/15