- Sat Feb 14, 2015 3:38 pm
- Forum: Calculus
- Topic: Given z=x²+3xy-2y²
- Replies:
**2** - Views:
**94**

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:
**398**

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:
**398**

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:
**398**

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:
**285**

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:
**394**

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:
**435**

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"???

- Sun Dec 14, 2014 11:34 pm
- Forum: Trigonometry
- Topic: trigonometric equation
- Replies:
**1** - Views:
**353**

Find All the value of x if 0=x equal/less than 2Pi that satisfy each of the following equation 1. cos x=1 This is just one of the basic cosine values. Have you done sine, cosine, etc, yet? You should know this from the graph of cosine. 2. sin x=1/2 This is another one of the reference values. You s...

- Fri Oct 31, 2014 6:28 pm
- Forum: Intermediate Algebra
- Topic: Qudratic systems???
- Replies:
**3** - Views:
**503**

K_V8 wrote:y - 1/2 * x^2 = 1 + 3x

y + 1/2 * x^2 = x

Does that help.

Not really. Is the LHS on the first equation (y - 1)/(2x^2) or y - (1/2)x^2 or something else?

K_V8 wrote:I keep getting 0 for x^2

Can't comment on your work until you show it. Sorry. Please reply showing all your steps. Thanks.

- Fri Oct 31, 2014 6:26 pm
- Forum: Intermediate Algebra
- Topic: quadratic eqaution using substitution?
- Replies:
**3** - Views:
**557**

x = y^2+2y+1 y = x-4 How Do I isolate y from the first equation.... Use the quadratic formula to solve for y in terms of x. But it would be much easier to follow the suggestion of the other poster: solve the second equation for x and set those equal. Do the instructions specify that you're not allo...