- Fri May 15, 2015 8:41 pm
- Forum: Intermediate Algebra
- Topic: Simplify the following expression...how?
- Replies:
**3** - Views:
**60**

Simplify the following expression. Write the answer with positive exponents and assume the variable represents a nonreal number. - 1/9 w^-8 times (-36w^-3) I don't see why you're supposed to "assume the variable represents a [complex or imaginary] number"...? Before the "times",...

- Fri May 15, 2015 7:02 pm
- Forum: Intermediate Algebra
- Topic: linear programming question - Problem with Constrain
- Replies:
**2** - Views:
**48**

I'm sorry but I don't understand what you're trying to do. What is the EXACT text of the exercise? What are the EXACT parameters? How are the constraints defined? What do you mean by "submitting as two scores"?

Thanks.

Thanks.

- Thu Apr 30, 2015 4:03 pm
- Forum: Calculus
- Topic: integral cos (Pi * x)
- Replies:
**1** - Views:
**105**

can anyone help me with the intigral of (pi * x) Since \,\pi\, is just a number, the integral of \,\pi\,x\, works just like the integral of any other product of \,x\, and a number: \int\, ax\, dx\, =\, a\, \left(\dfrac{x^2}{2}\right) In your subject line, you also ask about the integral of ...

- Sat Feb 14, 2015 3:38 pm
- Forum: Calculus
- Topic: Given z=x²+3xy-2y²
- Replies:
**2** - Views:
**335**

Given z=x²+3xy-2y² Estimate the change in z when x changes from 2 to 2,5 and y changes from 3 to 2,5 What method are you supposed to use to do the estimation? Did they give you any formulas, like partial-dz = (partial-dz/partial-dy)dy + (partial-dz/partial-dx)dx? Or maybe something like is in this ...

- Sat Jan 24, 2015 4:03 am
- Forum: Calculus
- Topic: Help needed to evaluate limits
- Replies:
**5** - Views:
**781**

Do you know how to read? Yes, I do. Do you? I showed you all the steps and explained that the result means that the limit of negative 2 times negative numbers, as those numbers get really, really far negative, is positive infinity, so the limit is "+infty". Why do you think showing your w...

- Fri Jan 23, 2015 8:07 pm
- Forum: Calculus
- Topic: Help needed to evaluate limits
- Replies:
**5** - Views:
**781**

Anonmeans, thank you, but can you please answer my question? I gave you all the steps. I showed you the limit expression is -2x. I pointed out that x < 0. Algebra says x < 0 means -2x > 0 so, when you take the limit, the value has to be positive. What part do you still need answered for you? :confu...

- Fri Jan 23, 2015 5:06 pm
- Forum: Calculus
- Topic: Help needed to evaluate limits
- Replies:
**5** - Views:
**781**

I have no idea how to solve it. Update! I just did this question: a) 3.5 B) infinity, but I am not sure why positive infinity??? Can someone explain it to me, please? The answers are right, I just need to find out why... How did you "do" these if you don't know how they're done? :confused...

- Fri Jan 02, 2015 5:19 am
- Forum: Trigonometry
- Topic: trig identities
- Replies:
**2** - Views:
**522**

I'm having difficulty solving the following (giving answers in degrees between -180 and 180): cos2(@) = sin(@)cos(@) The 2 is inside the cos, right? There's probably loads of ways to do this. One is to do the RHS as 2*(1/2)*sin(@)*cos(@) = (1/2)[2 sin(@) cos(@)] = (1/2) sin(2@). Then cross-multiply...

- Mon Dec 22, 2014 5:46 pm
- Forum: Trigonometry
- Topic: solving trigonometric eqn: 16sin^2theta = 3 / (cos^2theta)
- Replies:
**3** - Views:
**692**

I'll use "@" for "theta". Solve the following equation for theta: 16sin^2theta = 3 / (cos^2theta) I'd cross-multiply: 16sin^2(@)cos^2(@) = 3 Then use a trig ident: 2sin(@)cos(@) = sin(2@) 16sin^2(@)cos^2(@) = 4(4sin^2(@)cos^2(@)) = 4[(2sin(@)cos(@))^2] = 4sin^2(2@) Then: 4sin^2(2...

- Wed Dec 17, 2014 1:24 am
- Forum: Geometry
- Topic: Triangle Questions
- Replies:
**4** - Views:
**6149**

studentin wrote:Hi everyone i have two geometry questions can you please help me?

I can see part of one picture. What are the "two geometry questions"???