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For anyone who is willing and able to help me with some math questions, I’d greatly appreciate your assistance.
Also, I can assure you that this is not a homework assignment. The following math problems are merely sample questions taken from a sample math placement test that I am tyring to solve in order to better prepare myself for the actual test. Sadly, however, I have only been required to take 2 math courses so far(College Algebra and Prob. Stats), both of which were taken during my first 2 semesters at school - which was approximately 2 years ago. Needless to say, my math is rather rusty and I’m afraid I cannot even begin to solve these problems. Thus I humbly ask for your assistance.
Since writing this post Anonymous may have helped people, but has not within the last 4 days.
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-13 y and…teh second one…never seen anything like it. srry if I’m wrong on the first one.
1. You can solve a few different ways, depending on your preference. The most important thing to note is that the “-3″ is being multiplied by both “4y” *and* “-y”.
So the easier way of solving it would be to simplify inside the parentheses first,
2y-3(4y-y)=2y-3(3y), then apply the multiplication of -3 by 3y, getting: 2y-9y=-7y.
Another way of doing it would be to directly multiply both terms inside the parentheses by -3 first, 2y-12y+3y, (where -3*4y=-12y, and -3*-y=3y). Then solving 2y-12y+3y=-7y.
But the first way is easier.
2. (3x^3*y)(-2x^2*y^3) When you multiply two of the same number together that are each raised to a different power, such as 2^3 (8) and 2^2 (4), the answer is the number raised to the power of both added together. In my example, 2^(2+3)=2^5=32. Which is also the same as 2^3*2^2=4*8=32. All of the terms in the two parentheses are multiplied together, so you add the powers they’re raised to for each number, x and y. So you have 3*-2*x^(3+2)*y^(1+3)=-6*x^5*y^4
I like the first way for #1. Ah, it seems to be coming back to me now.
Here are a few more questions that are listed on the sample test. There are a total of 10 questions altogether. I’d like to see how each problem is solved if you don’t mind showing me. This is a relearning process for me ^_^
3. One factor of 18x^2-32 is
(The answer is 3x + 4, but I do not understand how to work the problem)
4. a/1+1/a =
5. 8/x^2-4 + 6/3x-6 =
No, I don’t mind =)
3) Factoring involves undoing polynomial multiplication. Such as turning x^2+2*x+1 into (x+1)*(x+1). It requires a little bit of guessing sometimes.
18x^2-32, you want to make it easier to work with so you can divide out by 2, and get 2*(9x^2-16). I’m going to ignore the 2* in the front for now, and put it back in later. Now you have to put it in some form of (_x+_)*(_x+_) In this case it would have to be (9x+_)*(x+_) or (3x+_)*(3x+_) to get the 9x^2 term. So you take a guess at one of the possibilities and work from there. For example, I’d start with (3x+_)*(3x+_). Let’s call the first number Z and the second number Y. (3x+Z)*(3x+Y)=9x^2-16. So in this case, 3x*Z+3x*Y=0x, since there are no normal x terms in 9x^2-16. So you would know that 3Z+3Y=0, or Y+Z=0, or Y=-Z. And Z*Y=-16. So looking at both of those, you’d get that Y is 4 and Z is -4, or the reverse. It doesn’t matter in this case since both sides have a (3x+_). So the result is 2*(3x+4)*(3x-4). Where the 2* is put back in, though it isn’t one of the factors.
There is also a formula for solving these, but it is more complicated, and at this level it’s normally just a bit of guesswork to find the answer instead since the answers are whole numbers. It’s called the quadratic formula though.
4) a/1+1/a. a/1=a. So it’s a+1/a. To add the two together, they need to be divided by the same number. As an example, to add 1 and 1/2 the hard way you would multiply 1 by 2/2, so you would get 2/2+1/2=3/2. It’s the same thing in this case. You multiply the a term by a/a (which is equal to 1, and multiplying by 1 doesn’t affect what an equation is equal to). So you’d get a*a/a+1/a=a^2/a+1/a. Then you can add the two together, getting (a^2+1)/a
5) 8/x^2-4 + 6/3x-6. Is it 8/(x^2-4) + 6/(3x-6)? If so, I’d look at the second term, 6/(3x-6) and simplify, by dividing out a 3. Getting 6/[3*(x-2)] = 2/(x-2). So overall it’s 8/(x^2-4) + 2/(x-2). You have to factor x^2-4, similar to problem 3) above, getting (x-2)*(x+2). So now it’s 8/[(x-2)*(x+2)] + 2/(x-2). You can’t quite add the two terms because they both need to be divided by (x-2)*(x+2). So you have to multiply the second term by (x+2)/(x+2) (equal to 1). And you’d get 8/[(x-2)*(x+2)] + 2*(x+2)/[(x-2)*(x+2)]. Now you can add the two together, getting (8+2x+4)/[(x-2)*(x+2)] = (12+2x)/[(x-2)*(x+2)] overall.
To give a little more explanation on factoring:
(5x+2)*(3x+4) = 5x*3x+5x*4+2*3x+2*8 = 15x^2+26x+8. There are four terms altogether, but you add two of them.
So in general, A*x^2+B*x+C being turned into (Lx+M)*(Nx+O), you know that L*N=A, M*O=C, and L*O+M*N=B. From there you just try different combinations.
So with a term such as 6x^2+5x-4, the -4 on the end shows you that either M or O is negative, but not both of them. So you know the factoring will be of the form:
(_x+_)*(_x-_). And either be (3x+_)*(2x-_) or (2x+_)*(3x-_) or (6x+_)*(x-_) or (x+_)*(6x-_) [In this case, it’s (2x-1)*(3x+4)]
If the second term was negative, such as 6x^2-11x+4, but the third term was positive, you would know that the answer is in the form (_x-_)*(_x-_). And if both terms are positive, then you know it has (_x+_)*(_x+_). In this example, it’s (2x-1)*(3x-4). Everything else is just guesswork, except when using the Quadratic Formula ^^
Now I remember why I disliked factoring - the guesswork involved.
I may have to practice some problems similar to #4. Actually, I fear that my greatest difficulty is going be determining which formula or solving procedure to apply to each problem. I have a tendency of attempting to solve one problem the same way for another because they appear similar. Hopefully I can avoid that this time.
Thank you kindly for all of your help thus far. I will post some new problems soon.
Factoring is a problem type all in itself.
But as for the others, you just have to look at how to align them into the right format to combine them together. Getting the denominators the same in fractions so you can add them together is the majority of those ones.
Here are 3 more:
6. xy+y/y * x^2/x^2+x =
7. Which line is parallel to the line having equation 2y+4x = 3?
A. y+2x = 3
B. y-2x = 3
C. 2y-4x = 3
D. -2y+4x = 3
E. 4y+2x = 3
8. If log3^x = 5, then
A. x = 15
B. x = 3^5
C. x = 5^3
D. x^3 = 5
E. x^5 = 3
Is this problem,
6) xy+y/y * x^2/x^2+x = [(xy+y)/y] * [(x^2)/(x^2+x)]?
I would start by simplifying each of the terms in parentheses. If you notice, in the first fraction, every term has y in it. And in the second, every term has x in it. So you can dividing it out. This is another way of writing the above:
[(xy/y+y/y)] * [(x^2)/x*(x+1)]. Because if you divided a number plus another number, it’s the same as dividing each individually by that number. And also, when you multiply two numbers added together by a number, it’s the same as multiplying each one individually. So you can say x^2+x = x*(x+1), and (xy+y)/y = xy/y+y/y.
Then, simplifying: [(xy/y+y/y)] * [(x^2)/x*(x+1)] = [(x+1)] * [x/(x+1)] = (x+1)*[x/(x+1)]. And the x+1 terms cancel, leaving x.
7. Which line is parallel to the line having equation 2y+4x = 3?
To be parallel it’d have to have the same slope as it. So the increase in y for every increase in x (the slope) has to be the same for both equations. In this case, you have to solve the equation for y. So you have 2y = 3-4x –> y = (3-4x)/2 –> y = 1.5-2x
So y has a slope of -2. (And a y-intercept of 1.5)
And out of these problems, A has the same slope. Because if you solve it for y, y = 3-2x, which is a slope of -2 (And a y-intercept of 3). So they’d both be parallel, (having the same slope), but with a different y-intercept so the A equation would be 1.5 higher than the other one.
8. log3(x) = 5 –> x=3^5. For any type of log base, when logN(y) = z, y = N^z.
Because log is basically the opposite of using an exponential. So for example, log base 10, log10 of 1000 is 3. log10(1000) = 3, because 10^3 = 1000. Basically log is seeing how many times you have to multiply the base to get the number you are taking the base of. Which is 3 in this case.
So just to verify, in the final step of #7 where y = (3-4x)/2, the 3 and -4x were both divided by 2 to arrive at y = 1.5 -2x, correct? Is it possible to graph the two lines in order to see if they are parallel?
I will have to study #6 better when I have more time. Thank you, DarkSnow. I will post the final 2 questions at another time.
That is correct. Multiplication and division are distributed to each term that is added or subtracted from each other.
Similarly, (6+4)/2=10/2=5, or you can use: (6+4)/2=6/2+4/2=3+2=5. And it would be the same with (14-4)*3, you could use 14*3-4*3=42-12=30 or do (14-4)*3=10*3=30.
It is possible to graph them. You start by putting a point at the offset, so y=1.5-2x would have a point where x=0 and y=1.5. Then go down two y for every one increase in x from that point (because of the -2x term, y is -2 for every 1 x), and plot the line. Then you could also graph the other line and visually see that they’re parallel.
If you’d like, you can try something like this, too:
You’re welcome =)
Sorry for the delay. I cannot access my other account at the moment.
At any rate, here are the final two questions:
9. For what value of K does the system of equations (3x + 2y = 5, 12x +ky = 9) have no solution?
10. tan(2 x π)/3 =
9) For no solution, the two lines would have to never cross. That would only happen if they were parallel. Rewriting them gives:
3x+2y=5 –> y=2.5-1.5x
12x+ky=9 –> y=9/k-12x/k
To have no solution and to be parallel, the slopes would have to be same. In other words, -1.5x=-12x/k –> -1.5*k=-12 –> k=-12/-1.5=8, So k would be 8.
10) This requires a knowledge of the tangent function. Tangent is sine over cosine. Or in other words, tan(2*π)=sin(2*π)/cos(2*π). sin(2*π)=0, so tan(2*π) = 0/cos(2*π) = 0/1 = 0. And by that, tan(2*π)/3 = 0/3 = 0.
Thank you for all of your help, DarkSnow. My apologies for the late response.
I didn’t forget :)
You’re welcome Mills. =) I wish you much success when you are studying these concepts and taking the test (?). ^^
You’re very kind. Thanks.
Fortunately, I still have a sufficient amount of time to prepare myself for the test. Someone was also nice enough to buy me a study guide and a book to facilitate the learning process.
Excellent. If you have further questions or want me to explain how I look at certain problems, feel free to ask =) Though I can’t promise too quick of replies. School is getting intense towards the end here, so I’m unreliable ;)
Ah, no worries. I completely understand that school comes first as it is of greater importance.
I may seek you assistance in the future should I need it.
Good luck in your last part of school this semester :)
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