Hi, I'm in need of some math help. I'll award points for people with answers.
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Hi, I'm in need of some math help. I'll award points for people with answers.

[From: ] [author: ] [Date: 12-08-13] [Hit: ]
but Im not sure which term it is...EDIT: You would use the binomial theorem. The term with no power of x would have xs that cancel out. I dont know if there is a more rigorous way to do this than simply testing the choices given by figuring out the coefficients for each term.......

3) I'm not exactly sure about this one; sorry. I think I know the answer, but don't know how to solve it. When expanding it, there is the constant term 84. This means the answer cannot be a), but I'm not sure which term it is...
EDIT: You would use the binomial theorem. The term with no power of x would have x's that cancel out. I don't know if there is a more rigorous way to do this than simply testing the choices given by figuring out the coefficients for each term. You know the answer cannot be a), as explained above. So test b) through e).
b) sixth: (1/x)^(9-5)(x^2)^5 = (1/x)^4(x^2)^5 = (1/x^4)(x^10) = x^10/x^4 = x^6 (No.)
c) fifth: (1/x)^(9-4)(x^2)^4 = (1/x)^5(x^2)^4 = (1/x^5)(x^8) = x^8/x^5 = x^3 (No.)
d) third: (1/x)^(9-2)(x^2)^2 = (1/x)^7(x^2)^2 = (1/x^7)(x^4) = x^4/x^7 = 1/x^3 (No.)
e) fourth: (1/x)^(9-3)(x^2)^3 = (1/x)^6(x^2)^3 = (1/x^6)(x^6) = x^6/x^6 = 1 (Yes.)
The answer is e) fourth.

4) f(a) = 2^a, log 2f(a)
I assume you mean base 2. I'll denote it with parentheses []:
log[2]f(a)
Substitute f(a).
log[2](2^a)
The answer is a. Raising the base (2) to the power of a yields the argument (2^a). Answer is c) a.

5) log[10](x+4) + log[10](x-4) = 1
Combine logarithms.
log[10](x+4)(x-4) = 1
Multiply.
log[10](x² - 16) = 1
Raise 10 to the power of both sides.
x² - 16 = 10
Add 16.
x² = 26
Take square root of both sides.
x = √26
Therefore the answer is b) radical 26.

6) There's really no way to answer this since we don't know what AB is...

ANSWERS:
1) b) -1/4 ≤ y ≤ 1/4
2) e) {2π/3, 4π/3}
3) e) fourth
4) c) a
5) b) radical 26
6) You didn't provide enough information to solve the problem. We don't have a diagram so we don't know what AB is.

Hope this helps :)

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I don't know, I'm not a doctor!!

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Hints:
1. sine varies between -1 and 1.
2. I'd replace sin^2x by 1 - cos^2x and solve the resulting quadratic for cos x.
3. All the terms are of the form constant * (1/x)^m * (x^2)^n where m + n = 9. You need those powers to cancel out. The first term has m = 0, the second term has m = 1, the 3rd term has m = 2, etc.
4. log2 cancels out 2 to the power.
5. log10(a) + log10(b) = log10(ab).
6. Draw the picture and see what you can figure out about the size of the box. It's obviously as high as the cylinder.

Now do your own homework/test.
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