Mt.SAC MATH HELP

what school do you go to?

Dead Man
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Mt.SAC MATH HELP

Post by Dead Man » Fri Jan 16, 2015 5:48 pm

By any chance does someone else go to this College.

I am currently attending but the Math Placement test is holding me back LOL.

I had taken it before but I never took a math class so now 2 years have passed since I took it and my scored expired :|

I already set my appointment but hopefully I get a good score and I can get a MATH class for this spring semester :|

I need to freshen on some practice question like these below. I haven't taken a math class like in over 5 years so I need to freshen up.
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Last edited by Dead Man on Fri Jan 16, 2015 7:59 pm, edited 1 time in total.
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SS Sparky
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Re: Mt.SAC

Post by SS Sparky » Fri Jan 16, 2015 6:54 pm

I CAN HELP :D But what do you need help with? You want answers? Or you want like how they're done?

Dead Man
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Re: Mt.SAC

Post by Dead Man » Fri Jan 16, 2015 7:07 pm

SS Sparky wrote:I CAN HELP :D But what do you need help with? You want answers? Or you want like how they're done?
Both if it is not to much trouble.
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SS Sparky
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Re: Mt.SAC

Post by SS Sparky » Fri Jan 16, 2015 7:51 pm

1. First you need to know 3 minus what gives -1. We know that's -4. So (2x+4)=4, but because it's l2x+4l, that means an answer of -4 would also pump out the number 4. So right now we have (2x+4)=4 AND -4. We solve that and get that x can be both 0 and -4. So the answer is D.

2. For this one it's just move everything to the other side. Since we're trying to find h, then we move everything that's not h to the other side, which gives S-2wL=2wh+2Lh. They both have an h on the right side so we can take them out which gives S-2wL=h(2w+2L). Move that big parenthesis on the right to the left and get (S-2wL)/(2w+2L)=h, so C.

3. First thing to know is that a negative exponent means that number is flipped from numerator to denominator and vice versa. So what we really have right now is ((x-y)^(-2))/((x+y)^(-1)). We said that they flip so by taking out the negatives we get (x+y)/(X^2-y^2) which is D.

4. Exponents are added when multiplied with other exponents, BUT if the exponent is having its own exponent, then you multiply. For example, x^3 times x^4 is x^7, BUT (x^3)^(4) is x^12.[/color] But if we follow what we said about exponents from number 3 we can do it like this. 3x^(1/5) remains that while we work on the other one. 8x^(-3) is 8/(x^3). Now it also has a -2/3 outside, which means now the 8/(x^3) is ((x^3)/8)^(2/3). Now we multiply that into 3x^(1/5). That becomes ((3x^(1/5)x^(3)^(2/3)) over (8^(2/3)). So the easy part is the bottom, which is 4 because 8^(2/3) is 4. The top would be 3x^(1/5)x^(2) which is 3x^(11/5). SO the answer is A.

5. First we see if this is FOILable. (x+1)(2x-3). From that equation it's FOILable. Iunno if you know FOIL, but FOIL means the second degree equation (second degree because the highest exponent is a 2) or even higher degrees can be separated into simpler equations like that. So right now we have (x+1)(2x-3)=0. We just need to know what x has to be to make their respective parenthesis ZEROES. x+1=0 and 2x-3=0. We find that x can be either -1 or 3/2. So the answer is C.

6. By changing the x in the f(x), it just means to plug in whatever is within those parenthesis into the variable. For example, f(SPARKYRULEZ)=x^(infinity) is (SPARKYRULEZ)^(infinity). What we have right now is a (b-1) so we plug in b-1 into the x. (b-1)^2-5(b-1)+1. For times sake, that equals to b^2-2b+1-5b+5+1. We simplify that to b^2-7b+7, which is A.

I have to go out for 30min but when I come back I'll finish it :D hope it's helping

Dead Man
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Re: Mt.SAC

Post by Dead Man » Fri Jan 16, 2015 7:59 pm

Wow thanks for taking the trouble of helping me and giving detail explanations it really helps.
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SS Sparky
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Re: Mt.SAC MATH HELP

Post by SS Sparky » Fri Jan 16, 2015 8:48 pm

Nah it's fine haha I just graduated, so I have a lot of free time on my hands :D

7. So let's learn how each of the choices should look like in equation form.
Parabolas only have x^2. So things that (once simplified) should have a base equation of x^2+b=y. You can have 27052305x^2 + 91874917 = 19275175. That's still a parabola. Anything can change, there just can't be anymore higher degree variables than 2. So ONLY x^2.
Circles have a base formula of x^2+y^2=1. Honestly as long as the two squares are there and it equals 1 then it's a circle.
Hyperbolas have a base formula of ((x^2)/(a^2))-((y^2)/(b^2))=1. The a and b denotes slopes and vertices of each opposing parabola formed. Don't worry about those now cuz all we care about is what the equation is.
Ellipses are practically the same as hyperbolas except the minus is a plus.
Finally we look back at our equation.I'm gonna cheat because the other way is too long :P We see that if we just purely remove EVERYTHING except for the x^2 and y^2, the correlation between those two is a minus, so it is x^2-y^2. Out of all the base equations I gave to you, the only one with a x^2-y^2 is hyperbolas, so the answer is C. You can try to simplify it, but it's gonna kill time which I'm sure you don't have.

8. For this, we try to cancel things out. Since we're trying to find the x, we try to cancel the y. To do this, we have to make the 2 ys from both equations cancel each other by being the negative form of the other. To do this we will multiply the first full equation by 2 to get 6x-2y=8. Now we can put the two equations on top of each other and cancel. 6x and 3x gives 9x. -2y and 2y gives 0. 8 and -5 gives 3. So the final equation is 9x=3 and x becomes 3, which is A.

9.Logs are stupid and retarded and I hate them but what must be done must be done. When logs add, that means they're multiplied. For example, log(xy)=logx+logy. SO for this we have logx+log(x+3) which means it's log(x(x+3))=1. What a log really means in this form is 10^1=x(x+3). If you ever see logs, sometimes they're log6(x)=1 (where the 6 is a subscript). That would mean 6^1=x, so x is 6. When there are no subscripts, it means a base of 10. So for our equation, it means 10^1=x(x+3). 10=x^2+3x or x^2+3x-10=0. We need to FOIL again, and we would get (x+5)(x-2)=0. We find that x=-5 and 2, so C.

10. This is just pretend the less than is an equal. (2x-5)(x+3)=0. x could be 5/2 or -3. Now we plug and chug. We plug in 3 numbers. One that is higher than our highest, one that is in the middle of the two numbers, and one that is lower than our lowest. I'm choosing 3, 0, and -4 because they're easy. Plugging in 3, we find that 6<0, which is false. Plugging in 0, we get that -15<0 which is true. Plugging in -4, we get 13<0, which is false. So now we know that any number above 5/2 and any number under -3 cannot be true. So, only numbers within those numbers are correct. Our choices are now B and C. Since we know that 5/2 and -3 would give us 0 which is not less than 0, we know that they cannot be the values, so the only answer is B.

11. Honestly, we can just plug in these two points into each equation and see if they work. If they work then that's the answer. BUT since we're learning :D we need to make it ourselves. Their base equation is a line with a formula of y=mx+b where m is a slope and b is a constant. To find slope we do (y2-y1)/(x2-x1), so for us it is (6-8)/(3-9)=2/6=1/3. So now we know our equation is y=1/3x+b. Now we pick one of the points and plug it in. Let's do (3,6). We get 6=1+b. b as we find is 5. So our full equation is y=1/3x+5, which is C.

Hope you had fun learning :P

ladyToOLs
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Re: Mt.SAC MATH HELP

Post by ladyToOLs » Sat Jan 17, 2015 9:06 pm

WOW sparky!! Thanks for the math refresher!

SS Sparky
Posts: 298
Joined: Fri Dec 06, 2013 11:53 am

Re: Mt.SAC MATH HELP

Post by SS Sparky » Sat Jan 17, 2015 10:55 pm

It's what I doooossssssss. Glad to help anyone in anything math or science related :P

Dead Man
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Re: Mt.SAC MATH HELP

Post by Dead Man » Mon Jan 26, 2015 7:24 pm

Im doing a Practice test right now but it seems that I have completely forgotten my MATH =(

Any steps to this?
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This one I just have a question on step 4
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These are some steps I found on how to get the answer
n+3 = Total Games Won
n+10 = Total Games Played
1/2 = 50%

Step 1: (n + 3)/(n + 10) = 1/2
Step 2: n + 3 = 1/2(n + 10)
Step 3: n + 3 = n/2 + 5
Step 4: n/2 = 2 -- Now right here since it is a dividing sign dos that mean to get that 4 it was multiplied? Or was the step done here addition?
Step 5 : n = 4
Last edited by Dead Man on Tue Jan 27, 2015 1:32 am, edited 2 times in total.
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Okonkwo
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Re: Mt.SAC MATH HELP

Post by Okonkwo » Mon Jan 26, 2015 8:52 pm

oh mans math hw
haven't done maths in a loooooong time but lemme try.

First one,
i just memorized how the parabola graphs look
(a) y = ax^2
a changes how wide the parabola opens up. as it approaches zero, the wider it gets
if a is negative then it would open downwards, if a is positive then it opens upwards
(b) y = x^2 - a and (c) y = x^2 + a
a determines the y coordinate of the origin
so for y = x^2 - a, the origin would be at (0,-a)
(d) y = (x-a)^2 and (e) y=(x+a)^2
the a in this case determines the x coordinate of the origin
so y=(x-a)^2, the origin would be at (a,0)
just pay attention to where a is positioned, in front of x, outside the square, or inside the square
when u get to circles, hyperbolics, etc., good luck lol

so the answer would be c i think lolol
PS the above probably sounds super ridiculous, sorry
if that doesn't work, u could always just try to plot it out with a equal to a certain number to see how it compares to the answer choices

second,
Step 1: (n + 3)/(n + 10) = 1/2
Step 2: n + 3 = 1/2(n + 10)
Step 3: n + 3 = n/2 + 5
Step 4: n + 3 - 5 = n/2 + 5 - 5
Step 5: n - 2 = n/2
Step 6: n - 2 - n = n/2 - n
Step 7: -2 = (-1/2)n
Step 8: -2(-2) = (-2)(-1/2)n
Step 8: 4 = n


hope this helped?
good luck

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