555 wrote:Solve. and check by answer

4(1-p)= 3(p-2)

First do the parentheses; the left side would be 4(1) + 4(-p) = 4 - 4p. Then solve like they show here: add 4p to both sides, add 6 to both sides, and then divide thru.

- Thu Dec 02, 2010 1:03 pm
- Forum: Pre-Algebra
- Topic: linear equation: solve 4(1 - p) = 3(p - 2), and check answer
- Replies:
**1** - Views:
**2071**

555 wrote:Solve. and check by answer

4(1-p)= 3(p-2)

First do the parentheses; the left side would be 4(1) + 4(-p) = 4 - 4p. Then solve like they show here: add 4p to both sides, add 6 to both sides, and then divide thru.

- Tue Sep 14, 2010 11:15 am
- Forum: Advanced Algebra ("pre-calculus")
- Topic: given u=-28.593*ln(k)+144.76, empirically define c, j
- Replies:
**1** - Views:
**1618**

u = -28.593*ln(k) + 144.76 The object is to empirically define "c" and "j" in the following equation by back substitution: u = c*ln(j/k) ; only valid where k approaches j You can use log rules to take the multiplier out front of the logarithmic term inside as an exponent. Then y...

- Mon Jun 07, 2010 12:18 am
- Forum: Pre-Algebra
- Topic: proportional or non proportional? 2Gb holds 250, 40Gb holds
- Replies:
**1** - Views:
**1551**

What did you get when you did the cross-multiplication and compared the results?

- Sat May 01, 2010 8:34 pm
- Forum: Intermediate Algebra
- Topic: absolute value quadratic inequalities
- Replies:
**1** - Views:
**2410**

Do the graph of the quadratic. Anywhere it's below the axis is negative, so re-draw that part upside-down so it's above the axis. Solve x^2-1=0 to find where the original graph crosses the axis and where the new graph touches. Where x^2-1 was positive anyway, solve x^2-1<2. Where x^2-1 was negative ...

- Mon Apr 12, 2010 10:48 pm
- Forum: Intermediate Algebra
- Topic: How do I determine the coordinates of the vertex ?
- Replies:
**1** - Views:
**990**

With the equation 4(x-2)(x+6) , how would I determine the coordinates of the vertex ? One way is to multiply out the two factors (not the 4) and then use the formula they give here . A "cheat" way is to remember that the vertex is midway between the two x-intercepts. So use the two factor...

- Tue Feb 23, 2010 12:41 pm
- Forum: Intermediate Algebra
- Topic: Help Adding Rational Expressions
- Replies:
**3** - Views:
**1527**

-2 + 4y + 6 ------- ------- ----- LCD = (y+3) (y-3) (y-3) y^2 - 9 (y-3)^2 3-y -2 + 4y + 6 ---------- ---------- ------- (y+3)(y-3) (y-3)(y-3) -1(3-y) The denominator 3 - y is not equal to -1(3 - y); it is equal to -y + 3 = -y - (-3) = -(y - 3). Then you can put the negative on the 6.

- Sun Jan 31, 2010 1:16 pm
- Forum: Beginning Algebra
- Topic: Finding the principal invested
- Replies:
**1** - Views:
**1518**

Use the set-up they show here: http://www.purplemath.com/modules/investmt.htm

- Fri Sep 18, 2009 1:40 pm
- Forum: Discrete Math
- Topic: are any sequences both arithmetic and geometric?
- Replies:
**1** - Views:
**2478**

Are there any sequences which are both arithmetic and geometric? I don't think so, but how would I prove this?

- Tue Sep 08, 2009 4:52 pm
- Forum: Uncategorized
- Topic: How many metres would the pony cover when taking 24 steps?
- Replies:
**1** - Views:
**1761**

A horse takes three steps to walk the same distance for which a pony takes four steps. Suppose that one step of the horse covers a half-metre. How many metres would the pony cover when taking twenty-four steps? I keep tying myself in knots, getting nowhere. Any ideas for an intelligent start on this...

- Mon Jul 27, 2009 6:32 pm
- Forum: Matrix (Linear) Algebra
- Topic: vet designs special diet for cat.... Find least-cost mix
- Replies:
**2** - Views:
**4032**