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anonymous
 one year ago
Can anyone help me with this precal please.
anonymous
 one year ago
Can anyone help me with this precal please.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's the question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's in the attachment

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which one(s) are you having trouble with?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0All of them, but any help would be great.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, the first one is: a. Evaluate the expression 2a−3b/2 when a = −5 and b = −3 This means to plug the given values into the expression, and simplify. According to order of operations, the division gets done *before* the subtraction, so the expression (when typeset) looks like this: \[ 2a  \frac{3b}{2}\] Now, what do you get when you plug in 5 for a and 3 for b?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442606735382:dw That's how it looks, and I plugged in a and b on the 2nd one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, that's a different problem. So, we need to figure out which is the problem they're asking! What you had written was 2a  3b/2, which is \[2a  \frac{3b}{2}\] What you put in your posted image is (2a3b)/2, which is \[\frac{2a3b}{2}\] They're different, and they'll give different answers.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, I didn't know that. It's the second one then, (2a3b)/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then you're off to a good start! Just start simplifying the numerator in \[\frac{2(5)  3(3)}{2}\] ... what do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442607214084:dw Is that right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You got it! Good! Done with the first one!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you have an image for the second one? I suspect what appears in my attachment isn't the actual problem you want...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, that's the actual problem, it's that exactly :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, then question (1b) is: Simplify: 8y − 15 − 3(7 − y) When you put this in regular math style, it looks like this: \[ 8y  15  3(7y)\] You need to do the multiplication first, so multiply out that last part, 3(7y). What does that give you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.015.89? or 21? If I did it right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We're on a different problem now ... there's no longer any a = 5 or b = 3 in sight! (That was just for problem 1a.) Now, just multiply this out: \[3(7y)\] Your answer will have a "y" in it!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You need to use a distributive law: a(b+c) = ab + ac Think: What's 3 times 7? (write it down) Then think: what's 3 times y? (write it down)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm curious .... where did you get the 15.89y from?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03 times 7 is 21 3 times y is 5.10 I got 15.89 by putting 3(7y) directly into the calculator... which is why I thought it was wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There's no NUMBER being plugged in for y in this problem. 3 times y is just 3y . (Remember: a negative times a negative is positive) This problem isn't a calculator problem; it's an algebra problem (working with variables).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh, okay. I was confused about that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here's where we are so far: \[\begin{align} 8y−15−3(7−y) &= 8y  15  21 + 3y \cr &= \text{ ??} \end{align}\] Now, you want to combine like terms. (Combine the number parts; combine the y terms.) What do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry my OpenStudy is acting up and being really slow. Yes, 11y33?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My computer just locked up! I had to close my whole browser and reopen and come back here? Might have been something going on at OpenStudy. I like the 11y part. What's 15  21?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, it's 36 then plus 3 is 39?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unless it's negative 15 then it's 6 plus 3 which is 9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Remember: 15  21 means (15) + (21). Move to the left 15; move to the left 21 more; where do you end up? (You were *close* before when you got 33; just goofed up the arithmetic a bit.)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0with a minus sign in front: 15  21 = 36. Now, put it all together: \[ 8y−15−3(7−y) = 8y−15−21+3y = 11y  36\] and that's it!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, got it! :) I added an extra 3 in my problem that's why I kept saying "plus 3" and yea, I forgot my minus sign. Thank you.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm just kind of having fun here today; I visited this site a few years ago, and didn't even know if it still existed. Clearly it does! It seems like a great resource for the math community. I may try to make regular appearances here! I've been working on a Precalculus course for years now, and I'm more than 2/3 of the way done: http://www.onemathematicalcat.org/Math/Precalculus_obj/tableOfContentsPreCalculus.htm Lots of the stuff I was helping you with today is from my Algebra course (they always do algebra review in PreCalc): http://www.onemathematicalcat.org/algebra_book/online_problems/table_of_contents.htm You might want to check them out to see if they can help you. There's lots and lots of lots of practice with immediate feedback, so you can check your answers. Well, I'd really love to stay and help you more right now, but I've got some other things I've got to do. Just repost these as separate questions, and I'm sure someone will help you quickly! Have a wonderful day!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, well thank you sooo much! You have a good day too!
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