July 24, 2014

Posts by Noether

Total # Posts: 34

The volume of a sphere with radius r is (4/3)(pi*r^3). Use that.

What's the proper way to reference a book from Project Gutenberg (MLA format)? It doesn't have all the information for the book format, but it was originally published as a book, so the web page format doesn't seem right. (If it matters, the book is Lewis Carroll&#...

When an antibiotic is used, some of the bacteria may be more able to survive it than others. The more-resistant bacteria survive to reproduce and pass their resistant genes to the next generation. The less-resistant bacteria don't get the chance to reproduce. Therefore, th...

Looks right to me.

That's probably because of significant figures. 15.0 m and 95.0 N both have three significant figures. Therefore, the answer should also have three significant figures. 1425 rounded to three sigfigs is 1430.


Not quite. In your first step, you aren't doing the same thing to both sides of the equation. 5x-3 = 11+12x You correctly recognized that you need to isolate the x terms on one side. However, if you subtract 12x from the right, you need to subtract 12x from the left, not a...

No, 40 is right. [8(6 - 1) - 1] - [2 - (5 - 2)] [8(5) - 1] - (2 - 3) 40 - 1 - (-1) 40

double check me please. math 012
(4)(5-w)/3 = -w (20-4w) / 3 = -w 20/3 - 4w/3 = -3w/3 20 - 4w = -3w 20 = w. Looks right to me.

math 012 double check me please
Your first two steps are fine. On the third, in which you go from 12x - 12 = 10x - 43 to -12 = -2x - 43 I think you mean to say you're subtracting 12x from each side. Adding 43 to each side leaves 31 = -2x. Divide each side by -2 to get x = (-31)/2. So, it looks like you w...

math 012
Your mistake here is when you go from 16y + 8 = 10 + 12y to 8 = -6 + 12y. You aren't subtracting the same thing from both sides; you're subtracting 16y from the left, but 16 from the right. Assuming you mean to subtract 16y, you should end up with 8 = 10 - 4y Subtract ...

There's not enough information here to answer those. There are infinitely many functions that sent 9 to 2. For example, it could be f(x)=(x/3) - 1, f(x) = (x-1)/4, f(x) = x-7, f(x) = x^2 - 79, f^x = x^(1/2) - 1, f(x) = 2x-16, or any of infinitely many others. We need more ...

That's correct. I think it might be better if you made it clearer in your explanation that you did 70 + (7*45); as it stands now, somebody reading the explanation might think you meant you did (70+7) * 45.

Oops. Switch l and w in that last line.

A = l * w. A = 1050 ft^2 w = 15 ft. w = A / l = (1050 ft^2) / (15 ft) = ?

The key fact here is that (at least on a plane) the angles of a triangle add to 180 degrees. So we have three angles, of measures x, x, and x+15. We can set up the equation 3x+15 = 180. Solve for x, and figure out which of the three angles is the largest (If figuring out which...

Note: This answer assumes the expression is parenthesized as (x^2 - 4) / (x^2 - 4x + 4). Either you made a mistake, I did, or the book did. The graph of that expression doesn't have a hole. It does, however, have a vertical asymptote at x=2. Are you sure you're suppose...

Those actually both use the same method, the one you described for the second example. x=(-1) makes (x+1) equal to zero, and x=1 makes (x-1) equal to zero.

Your answers aren't right. You can check factorizations pretty easily by just multiplying. (3w)(3 - w^2) = (3w)(3) - (3w)(w^2) =9w - 3w^2 ((m-n)^2)((m-n)^2) = (m-n)^4 =/= m^4 - n^4. Some hints: 9w - w^3 : Factor out the w first. This leaves a difference of two squares. Rem...

college algebra
You're done. 9, 18, and 8 have no common factor other than 1. (9 = 3^2, 18 = 2*3^2, 8 = 2^3)

Remember, the diameter is twice the radius. (Equivalently, the radius is 1/2 the diameter)

To Everyone at Jiskha
Merry Christmas, Happy Hanukkah, Happy New Year, um . . . Io Saturnalia?

Not if it's a function in x. Think about what the y-intercept is - it's the value of f(x) for x=0. You can't get two different values out of that.

Remember that a^(-x) = 1/(a^x), as long as a isn't zero. So 10^(-4) = 1/(10^4) = something you should be able to figure out.

pre cal
For the first one, remember that (a^b)^c = a^(bc). 2/3 * 3/2 = 1, so raise each side to the power of 3/2. That's all you need to do; you'll be left with x=(something). For the second, start by splitting the left side into two fractions, like so: (2^x)/(2^x) + 4/(2^x) =...

This is a quadratic, so we'll try to factor it into (x+a)(x+b). The constant term is positive, so we know that a and b have the same sign. The x term is negative, so we know a and b are both negative. We need to find two negative numbers, a and b, such that a*b = 35 and a+...

math B
There should be a way to solve it, but we can't help without the diagram. We need to know where all the points are in relation to each other.

math B
This can't be answered without the diagram that should be with it.

What's 0.25% as a decimal? Once you know that, it's just multiplication. (Remember, 1% = 0.01)

Can you give an example of a problem you can't figure out? It'll be much easier to help with a specific example of the kind of thing you're having trouble with.

Both are grammatically correct (assuming you mean twenty-nine). There are guidelines some people like to follow for when to use the digits and when to write it out, but you should be fine using whichever seems to work best.

x^3 and 1x^3 are the same thing. The coefficient is just usually left off when it's a 1. As for the second problem, your answer isn't quite right. Recheck your x^3 and x terms; there should also be no constant term.

Use Newton's law of gravitation. g=(m_1)(m_2)(G)/(d^2) m_1 = mass of first body m_2 = mass of second body G = gravitational constant (value is unimportant for this particular problem) d = distance between bodies.

Intro to Computer Programming - Pseudocode
You don't need a 2D array. You probably do need a loop (although some languages have features that kind of let you avoid that). You'll need at least two 1D arrays. (Possibly 3, depending on how you implement it) What you need to do is this: Get the student's answer...

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