precalculus?

Question

i have a few questions that i dont understand how to do:
1. solve for 'a': V=2(ab+bc+ac)
i dont understnad how you would solve for the a
2. factor x^6-16x^4
3.rationalize the bottom: 2/(sq.rt.2)+(sq.rt.3)
4. solve: x^2-2x-8/x^2-7x+12
5.solve: 2x/4(pi) +1-x/2

if you could explain how to do these it would be very much appreciated

Answer 1

Hi,

1. solve for 'a': V=2(ab+bc+ac)
Distribute the 2 and eliminate parentheses.

V=2ab + 2bc + 2ac

Move all terms with "a" to one side and all terms without "a" to the other side.

V - 2bc =2ab + 2ac

Factor out a on the right side. (Do not factor out the GCF. You are trying to get just "a" alone.)

V - 2bc = a(2b + 2c)

Now divide both sides by the expression 2b + 2c. It will cancel out on the right side, leaving:

V - 2bc
------------ = a
2b + 2c


2. Factor x^6-16x^4

Take out the GCF.

x^6-16x^4
x^4( x^2 - 16)

The expression in the parentheses is the difference of squares, so it factors again:

x^4( x^2 - 16)
x^4( x - 4)(x + 4)

3.Rationalize the bottom: 2/(sq.rt.2)+(sq.rt.3)

........2
---------------
..__.......__
√2...+..√3

Multiply top and bottom times the conjugate of the bottom. The conjugate of the bottom is the ame expression with the sign in the center changed. So multiply top and bottom by
( √2 - √3 )

..............__.......__
........2(√2...-..√3...)
----------------------------------- =
....__.......__......__......__
(√2...+..√3...)(√2...-..√3...) FOIL the bottom. You get 2 - 3 = -1

............__.........__
........2√2...-..2√3
-------------------------
..............-1

This divides and equals:

.......__.........__
...2√3...-..2√2



4. solve: x^2-2x-8/x^2-7x+12

x^2.-.2x.-.8
------------------ =
x^2.-.7x.+.12

(x.-.4)(x.+.2)
---------------- =
(x.-.4)(x.-.3)

x.+.2
-------
x.-.3



5. Solve: 2x/4(pi) +1-x/2

I'm not sure about this problem. I can get a common denominator and add them, but there's really nothing to "solve".

2x........1.-.x
-----..+..-------
4pi..........2

The first fraction reduces:

x........1.-.x
-----..+..-------
2pi..........2

The common denominator is 2pi, so multiply 2nd fraction by pi, top and bottom.

x..........pi(1.-.x)
-----..+..-----------
2pi..........2pi

This simplifies and combines to:

x.+.pi.-.pi*x
------------------
.........2pi


I hope that helps!! :-)

Answer 2

1. V/2 = a(b + c) + bc
a = (V - 2bc)/2(b + c)

2. x^6 - 16x^4
= x^4(x^2 - 16)
= x^4(x + 4)(x - 4)

3. (sqrt(2) + sqrt(3))(sqrt(3) - sqrt(2)) = 3 - 2 = 1
So you have 2(sqrt(3) - sqrt(2))

Answer 3

V = 2ab + 2bc + 2ac
V - 2bc = 2ab + 2ac
V - 2bc = 2a(b + c)
(V - 2bc)/(b + c) = 2a
(V - 2bc)/(2 * (b + c)) = a

x^6 - 16x^4 =
x^4(x^2 - 16) =
x^4(x + 4)(x - 4)

2/(sqrt(2) + sqrt(3)) =
2(sqrt(2) - sqrt(3))/(2 - 3) =
2(sqrt(2) - sqrt(3)) / -1=
-2(sqrt(2) - sqrt(3))

(x - 4)(x + 2)/(x - 4)(x - 3) =
(x + 2)/(x - 3)

Answer 4

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