For these four practice problems, I had little idea what the hell to do (plugging in seemed worthless), could anyone help? Sorry if this isn't what the board is suppose to be used for. Answers in spoiler tags.

If k and h are constants and x^{2}+kx+7 is equivalent to (x+1)(x+h), what is the value of k?
Answer:

In the figure above, if the legs of the triangle ABC are parallel to the axes, which of the following could be the lengths of the sides of triangle ABC?

1) Multiply out (x+1)(x+h) by FOILing and equate it to x^{2}+kx+7. What's the first conclusion you must make? What's the second conclusion?

2) The most important thing here is the slope of that line. Take into account that the slope of the line is rise/run. What's a possibity for the value of rise? run? (looking at the answers) Once you have those two values, how do you find the hypotenuse?

3) Judging from the answer, it looks like you mean (1/2)f(sqrt t)= 4, not 1/[2f(sqrt t)]= 4. The use of parentheses here is very important. If f(x) = 2x - 1, what's f(sqrt(t))? Once you have this, plug it into (1/2)f(sqrt t)= 4 and solve for t.

4) Something is missing from this problem, but I think what you meant to say is define an odd integer (call it j) as j = 2k+1, where k is a positive integer. If this is the case, how do we represent twice the odd integer?

Multiply out (x+1)(x+h) and the answer is obvious.

There are two conditions here. First, the ratio of the two legs has to equal 0.4, because the line goes through the origin and (4, 10). The second condition is the Pythagorean theorem.
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This one CAN be solved just by plugging in: plug the definition of f into the equation, then solve for t.

For any integer k, 2k is even, and 2k+1 is odd. To be twice the value of an odd integer, just double that.

You may be overthinking this a little. Just LOOK at the expressions

x^{2} + (h+1)x + h

x^{2} + kx + 7

Don't they look similar to you? What would it take to make them look identical? (And you know the answer, too -- try plugging that in and see what happens.)

It might help if you know that for two polynomials (of equal degree), for the two polynomials to be equal for every x, each of the coefficients must be equal for every x. Your two polynomials are x^{2}+kx+7 and x^{2}+(h+1)x+h. Setting these two equal, with x^{2}+kx+7=x^{2}+(h+1)x+h, we can equate the coefficents of every power of x:

1=1 (the two coefficients of x^{2})
k=h+1 (the two coefficients of x)
7=h (the two coefficients of the constant term x^{0})