# Thread: I have an upcoming test, and could use some help with this study guide

1. ## I have an upcoming test, and could use some help with this study guide

I'm in danger of failing this course for the second time, and my super-awesome teacher sent me this study guide. He said if I can get the problems on this guides, I'll be able to pass the test. The problem is that I need things explained to me a bajillion ways from sunday before I can understand them.

If anyone can run through even a few of these, really in depth, that would be just the most amazing thing ever. More important than the problems themselves, though, is the theory behind them. I'm basically incapable of abstract thinking (I'm exaggerating, but only slightly), and any help in understanding the reason behind the problems, or what they mean, that'd also be super-helpful.

Thanks for any help anyone can give.

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(a) If y varies directly with x2 and y = 256 when x = 8, find an equation to represent this relationship.

(b) Find y if x = 2.

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(a) If y is inversely proportional to 6
 x

and y = 25
when x = 8,
find an equation to represent this relationship. (Round the constant of proportionality to three decimal places.)

(b) Find y if x = 4.
(Round your answer to two decimal places.)

--

Use the graph to answer the questions.

Consider the graph of f(x).

(b) Estimatef(0).

(c) Estimate x such that f(x) = 6.
(Enter your answers as a comma-separated list.

--

(b) Estimate h(9).
h(9) ≈

(c) Estimate h(−5).

h(−5) ≈

(d) Estimate x such that h(x) = −9.
(Enter your answers as a comma-separated list.)
x

--

Simplify the rational expression.
 5b3 − 4b2 − 9b 4b2

--

Simplify the rational expression.
 x4y2 − 6x3y4 + 9xy3
------------------- (this is an over line, I was having trouble copying it, the other site is being weird)
6
x
2
y
2

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Multiplying rationals

x-6 6x-7
-- * --
x-7 6-x

--
 w + 5 35w + 42
÷
 11w + 55 5w + 6

2. ## Re: I have an upcoming test, and could use some help with this study guide

I don't see that any of these involve "abstract thinking". They have everything to do with "knowing the definitions".

(a) If y varies directly with $\displaystyle x^2$ and y = 256 when x = 8, find an equation to represent this relationship.

(b) Find y if x = 2.
Easy if you know what "varies directly means. If "b varies directly with a", then b=ka for some number, k. If "y varies directly with $\displaystyle x^2$" then $\displaystyle y= kx^2$ for some number, k. If y= 256 when x= 8, then $\displaystyle 256= k(8^2)$. So what is k? Once you know k, what is $\displaystyle y= k(2^2)$?

You don't actually need to find k directly to answer this. 2 is 1/4 of 8 so y is 1/16 of 256. Do you see why 1/16 which is $\displaystyle (1/4)^2$?

If y is inversely proportional to 6 x

and y = 25
when x = 8,
find an equation to represent this relationship. (Round the constant of proportionality to three decimal places.)

(b) Find y if x = 4.
(Round your answer to two decimal places.)
Again, this is a matter of knowing the definition. "b is inversely proportional to a" means $\displaystyle b= \frac{k}{a}$. (1/a is the "multiplicative inverse" of a which is the reason for the term "inverse" proportional". ) This is also the same as $\displaystyle ab= k$ (multiply on both sides by a). Sinces "y = 25 when x = 8", $\displaystyle 25= \frac{k}{8}$ or, equivalently, $\displaystyle 25(8)= k$ so you know what k is. You can then find y when x= 4.
Or you could argue that, going from x= 8 to 4, x is divided by 2 so, because this is an inverse proportion, y must be multipliedby 2.

The next couple of problems give graphs and ask you to find y for a given x and x for a given y. All that is required here is knowing what a graph is and what information it is giving you. The vertical axis tells you "y", the height above (for positive y) or below (for negative y) the horizontal axis, and the horizontal axis tells you the distance right (for positive x) or left (for negative x) of the vertical axis.

The first asks you do "estimate f(0)". Okay, that graph is of y= f(x) so x is "right or left" and y is "up or down". x= 0 is exactly on the vertical axis. What is the y value there (how high above the horizontal axis)?