Need help and a memory refresher on 8 problems.. Please do not give me answers. I want to remember how to solve them. This is my midterm review.

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Need help and a memory refresher on 8 problems.. Please do not give me answers. I want to remember how to solve them. This is my midterm review.

Mathematics
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11. Perform the indicated operations and write in standard form. (-10 + square root -9) ^2
16. Solve the problem. A rain gutter is made from sheets of aluminum that are 24 inches wide. The edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 58 square inches. There are two solutions to this problem. Round to the nearest tenth of an inch.
21. Solve the linear inequality. Other than not solution sign, use interval notation to express the solution set and graph the solution set on a number line. x-1 over 16 >/ x-4 over 20 + 1 over 80

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Please help me finish. I still have a chapter test and midterm to do for another class. Im majorly behind since the system keeps glitching.
Hopefully in your lesson somewhere you learned that \[\Large i = \sqrt{-1}\] ? or no?
Yes I did last term.
ok great
what we can do is rewrite -9 as -1*9 and simplify like this \[\Large \sqrt{-9} = \sqrt{-1*9}\] \[\Large \sqrt{-9} = \sqrt{-1}*\sqrt{9}\] \[\Large \sqrt{-9} = i*3\] \[\Large \sqrt{-9} = 3i\] make sense?
Yes!
\[\Large \left(-10+\sqrt{-9}\right)^2\] turns into \[\Large \left(-10+3i\right)^2\]
Yep I have that.
Next you'll use the box method or FOIL or the distributive property to expand out \[\Large \left(-10+3i\right)^2\]
personally I like the box method but you can choose which one you want to do
So.. (-10+3i)(-10+3i)
yes
keep going
I got.. 100-51i
incorrect
Yeah...
Not sure where im going wrong.. (-10+3i)(-10+3i) 100-30i-30i+9i 100-60i+9i 100-51i
3i times 3i = 9i^2 not just 9i
since i = sqrt(-1), squaring both sides gives i^2 = -1
Omg -.-
91-60i
perfect
Okay. For number 16. Solve the problem. A rain gutter is made from sheets of aluminum that are 24 inches wide. The edges are turned up to form right angles. Determine the depth of the gutter that will allow a cross-sectional area of 58 square inches. There are two solutions to this problem. Round to the nearest tenth of an inch. I got to -20+- square root 296 all over -4
the square root 296 can turn into 2 square root 74
-20 +- 2 square root 74 all over -4 Then I can -20 and -4 --> 5
I'm kinda stuck there.
I'm not sure how you got that. What quadratic equation did you get to?
Um.. -2x^2 +20x -13=0 -20+- square root (20)^2 -4(-2)(-13) all over 2(-2) Some girl earlier said to- 2x+y=20 y=20-2x x*y=13 x*(20-2x)=13
Im really not sure. I followed what she said. And I responded saying I was stuck and she didnt reply.
ok let's start with a 24 inch line |dw:1444423527504:dw|
we will bend this line into a partial rectangle like this |dw:1444423553539:dw| I can't do animations on here but hopefully you can picture what is going on
I can.
after the bending is done, we have this left over |dw:1444423623561:dw|
if we make x = depth, then we have this |dw:1444423774023:dw|
what is the length of the bottom side?
What do you mean? As in y?
hint: |dw:1444423922673:dw|
replace the question marks with some expression in terms of x
I understand what your saying but I keep thinking Y. Unless its 58?
|dw:1444424117414:dw|
Or is it still 24 even though the edges have been bent.
Oh..
there are 2 xs being subtracted off, so that's how I got 24-2x
|dw:1444424187334:dw|
Yes I understand now.
Area = Length * Width 58 = (24 - 2x) * x 58 = 24x - 2x^2 -2x^2 + 24x - 58 = 0 solve for x using the quadratic formula
Okay. Im at.. -24 +- 4 square root 65 over -4
you made an error somewhere
-24 +- square root 576 +464 over -4
it should be -464 not +464
b^2 - 4ac (24)^2 - 4*(-2)*(-58) 576 - 464
Oops I put a + in front of the 4...
Now I have.. 6+- -1 square root 7
good, \[\Large x = 6 \pm \sqrt{7}\]
Answer-> 3.4 and 8.6
both are correct
Okay thank you.. 6 left.. 21. Solve the linear inequality. Other than not solution sign, use interval notation to express the solution set and graph the solution set on a number line. x-1 over 16 >/ x-4 over 20 + 1 over 80
This? \[\Large \frac{x-1}{16} \ge \frac{x-4}{20} + \frac{1}{80}\]
I know the answer is -10.. Its the only option.. My first attempt I canceled out the fractions by using 4... Yes thats it.
I would use 80 instead
True..
Multiply every term by the LCD 80 to clear out the fractions \[\Large \frac{x-1}{16} \ge \frac{x-4}{20} + \frac{1}{80}\] \[\Large 80*\frac{x-1}{16} \ge 80*\frac{x-4}{20} + 80*\frac{1}{80}\] \[\Large 5(x-1) \ge 4(x-4) + 1\]
Yep.. Should of figured that out :/ ... x
correct
Its [-10, infinity) correct?
yes, sorry I misread your first inequality it should be \(\Large x \ge -10\) which turns into `[-10, infinity)`
Okay. Sorry typed the wrong one.
Next problem.. 24. Determine whether the equation defines y as a function of x. xy + 5y = 1
I don't remember doing any problems like this. And I couldn't find any examples from my notes.
We solve for y to get xy + 5y = 1 y(x + 5) = 1 y = 1/(x + 5) which is definitely a function. You can graph to verify that it passes the vertical line test. Any given input (x) produces EXACTLY ONE output (y)
Okay!
26. Evaluate the function at the given value of the independent variable and simplify. f(x) = 3x2 - 4x + 4; f(x - 1)
I'm not sure on this one either.. I was thinking plug (x-1) in for the x's but im not sure. If you want me to post in a new question I can.
You're right. Replace each x with x-1 and simplify \[\Large f(x) = 3x^2 - 4x + 4\] \[\Large f(\color{red}{x}) = 3(\color{red}{x})^2 - 4(\color{red}{x}) + 4\] \[\Large f(\color{red}{x-1}) = 3(\color{red}{x-1})^2 - 4(\color{red}{x-1}) + 4\] now simplify the right hand side
` If you want me to post in a new question I can. ` yes please. It's getting laggy with all of these questions. It's a good idea to have one question per post to avoid clutter and lag
Okay Ill do it now..

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