# Thread: I'm confused with some high school level algebra, help please?

1. ## I'm confused with some high school level algebra, help please?

I have the answers, but I'm just not sure how to get to that. I'm hoping some one can explain to me.

1)Three people who work full time are to work together on a project, but their total time on the project is to be equivalent to that of only one person working full time. If one of the people is budgeted for _ of his time to the project and a second person for 1/3 of her time, what part of the third worker’s time should be budgeted to this project?

-answer is 1/6. I dont understand this either. I thought it was 1/3. since 3 people using 1/3 each of their time would equal 1 full time? I have no idea why there is a "_" in the question.

2)Simplify: 1/-2^-2

-answer: is -4. since the denominator has the power of a negative it rids of the numerator's existance. So what is left is just -2^2.... right? making 4 not - 4, since two negatives is positive... yes? no?

3)solve for x if: 1/x + 2/y = 3/z

so I know what I do in this is isolate the 1/x from the rest and some how get rid of the numerator of 1, but I'm not sure how to do that. Or if thats even the correct method.

4)If ((3/2)x) + 4 = 10, find the value of ((5/8)x) - 2

-answer is -7/6. in the first one i solved for x which turns out to be 4 and then I replace 4 into the x of the second equation which turns out to be 1/2. and it turns out I was complete wrong.

I can't do most of the questions in the last page. Never took pre cal.

2. Originally Posted by celestine
I have the answers, but I'm just not sure how to get to that. I'm hoping some one can explain to me.

2)Simplify: 1/-2^-2

-answer: is -4. since the denominator has the power of a negative it rids of the numerator's existance. So what is left is just -2^2.... right? making 4 not - 4, since two negatives is positive... yes? no?

I can't do most of the questions in the last page. Never took pre cal.
I'll take #2.

Use this rule of exponents:

$\displaystyle \frac{1}{a^{-m}}=a^m$

$\displaystyle \frac{1}{-2^{-2}}=-2^2=-4$

Keep in mind that $\displaystyle -2^2\neq(-2)^2$

Grouping makes a difference here.

3. that makes perfect sense!! thanks!

4. Originally Posted by celestine
I have the answers, but I'm just not sure how to get to that. I'm hoping some one can explain to me.

3)solve for x if: 1/x + 2/y = 3/z

so I know what I do in this is isolate the 1/x from the rest and some how get rid of the numerator of 1, but I'm not sure how to do that. Or if thats even the correct method.
Ok. I'll help on #3 as well.

$\displaystyle \frac{1}{x}+\frac{2}{y}=\frac{3}{z}$

Multiply all three terms by the common denominator $\displaystyle xyz$

$\displaystyle yz+2xz=3xy$

Isolate x terms to the left side of the equation:

$\displaystyle 2xz-3xy=-yz$

Factor out the x:

$\displaystyle x(2z-3y)=-yz$

Divide both sides by $\displaystyle 2z-3y$

$\displaystyle x=\frac{-yz}{2z-3y}\Longrightarrow\frac{-zy}{2z-3y}\Longrightarrow\frac{zy}{3y-2z}$

5. Originally Posted by celestine
I have the answers, but I'm just not sure how to get to that. I'm hoping some one can explain to me.

4)If ((3/2)x) + 4 = 10, find the value of ((5/8)x) - 2

-answer is -7/6. in the first one i solved for x which turns out to be 4 and then I replace 4 into the x of the second equation which turns out to be 1/2. and it turns out I was complete wrong.

6. Maybe its a mistake from the writers? Haha
maybe.

7. 1)Three people who work full time are to work together on a project, but their total time on the project is to be equivalent to that of only one person working full time. If one of the people is budgeted for _ of his time to the project and a second person for 1/3 of her time, what part of the third worker’s time should be budgeted to this project?

-answer is 1/6. I dont understand this either. I thought it was 1/3. since 3 people using 1/3 each of their time would equal 1 full time? I have no idea why there is a "_" in the question.

I don't understand it either.

But with the book answer given, (1/6 for the 3rd person), we can play backwards to determine that "-" for the 1st person.

"-" +1/3 +1/6 = 1
"-" = 1 -1/3 -1/6 = 3/6 = 1/2

So, did the author of the problem forget to write 1/2 in lieu of that "-"?

8. Here are some answers from your last page. Not much detail, but maybe it helps. Maybe some of the other guys here can finish up for you. Good luck.

1. If $\displaystyle f(x) = x^3 + 4$, then compute $\displaystyle f^{-1}(5)$

$\displaystyle f^{-1}(x)$ means the inverse of $\displaystyle f(x)$

To find the inverse function, let the x and y change places. Then solve for y.

$\displaystyle f^{-1}(x)\rightarrow x=y^3+4$

$\displaystyle f^{-1}(x)\rightarrow y^3=x-4$

$\displaystyle f^{-1}(x)\rightarrow y=\sqrt[3]{x-4}$

Now find

$\displaystyle f^{-1}(5)=\sqrt[3]{5-4}=\sqrt[3]{1}=1$

2. Find the domain and range of $\displaystyle f(x) = \sqrt{x^2-25}$

$\displaystyle x^2-25\ge0$ over the Real number domain.

$\displaystyle D=\{x|x\leq -5 \ \ or \ \ x\geq 5\}$ or $\displaystyle D=(-\infty, -5] \cup [5, +\infty)$

$\displaystyle R=\{f(x)|f(x) \geq 0\}$

3. If $\displaystyle f(x) = 2x^2 + 4$, find $\displaystyle f(x^4 – 1)$

$\displaystyle f(x^4-1)=2(x^4-1)^2+4\Longrightarrow2(x^8-2x^4+1)+4\Longrightarrow2x^8-4x^4+6$

4. If $\displaystyle z_1 = 3 + 2i$ and $\displaystyle z_2 = 4-5i$, find

a. $\displaystyle (z_1)(z_2)$

$\displaystyle (3+2i)(4-5i)=12-7i+10=22-7i$

b. $\displaystyle \frac{z_1}{z_2}$

$\displaystyle \frac{3+2i}{4-5i}$

Multiply both numerator and denominator by the conjugate of $\displaystyle 4-5i$

$\displaystyle \frac{3+2i}{4-5i}\cdot\frac{4+5i}{4+5i}=\frac{12+23i-10}{16+25}=\frac{2+23i}{41}$

5. Find $\displaystyle 8^{-\frac{2}{3}}$

$\displaystyle 8^{-\frac{2}{3}}=\frac{1}{8^{\frac{2}{3}}}=\frac{1}{\l eft(\sqrt[3]{8}^2\right)}=\frac{1}{4}$

9. Masters:
Wow thank you! Your huge post really helped me!
I looked through it, and I never knew that when you get an -Yi^2 that the Y will become positive...or negative depending on the situation.

ticbol:
Well, I don't think they would give me the answer key for on the test to work backwards, I wish though! Thanks for your insights, it was quite helpful regardless.