1. ## Clueless

Can someone give me a hand with these? I'm at a total loss.

View problems here.

Then a word problem (these always throw me in a loop).

Then a word problem too: "The height of a triangle is 4 inches less than the base. The area is 30 square inches. Find the height."

2. Originally Posted by SailorCardKnight
Can someone give me a hand with these? I'm at a total loss.

View problems here.

Then a word problem (these always throw me in a loop).

Then a word problem too: "The height of a triangle is 4 inches less than the base. The area is 30 square inches. Find the height."
For the word problem, note:

The area of a triangle is 1/2 x length of the base x height.
Call the length of the base "x". What is the height relative to this x?
Then you can use the "1/2 x length of the base x height" to solve for x...

3. Originally Posted by SailorCardKnight
Can someone give me a hand with these? I'm at a total loss.

View problems here.

Then a word problem (these always throw me in a loop).

Then a word problem too: "The height of a triangle is 4 inches less than the base. The area is 30 square inches. Find the height."
Hi SailorCardKnight,

$\frac{a^2-16}{4a} \cdot \frac{8a}{a-4}$

First, factor the numerator of the first fraction:

$\frac{(a-4)(a+4)}{4a} \cdot \frac{8a}{a-4}$

Multiply, then simplify:

$\frac{(a-4)(a+4)(8a)}{(a-4)(4a)}$

$\boxed{2(a+4)}$

Your second problem is a little difficult to read. I'll venture a guess:

$5-\frac{2}{5}x=\frac{7}{3}$

If this is not exactly it, you can still use it as a model.

First, multiply by the least common multiple of the divisors to clear the fractions.

$15(5)-15\left(\frac{2}{5}x\right)=15\left(\frac{7}{3}\ri ght)$

$75-6x=35$

Subtract 75 from both sides and then divide by -6 and you're done.

$y=-\frac{3}{4}x-17$

This is already in slope-intercept form, so we can easily tell that the slope is $-\frac{3}{4}$

Now, you want the slope of a line that is perpendicular to this one, so we must use the negative reciprocal of this slope.

The negative reciprocal of $-\frac{3}{4}$ is $\frac{4}{3}$

Using point (-1, 4) and our newly acquired slope of $\frac{3}{4}$, we can use the point slope form of the linear equation to find our required equation.

$y-y_1=m(x-x_1)$

It's just a matter of substitution now.

4. You guessed right on that one problem masters, thats exactly how I have it wrote. (I will admit my handwriting is awful...)