College Algebra problems..

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June 15th 2006, 09:17 PM
daniel4616

College Algebra problems..

Hey, I have some question on college algebra
1. Is the line y=|x| symetric about the line y=x

2. How do you prove points A(-1,2) B(3,-2) C(-6,-3) are vertices of a right triangle?

3. given f(x)= 2x if 0<x<1, 1-x if 1=<x<2 , 0 if 1=<x=<3: find the domain,f(1), f(2), f(3), and f(0.1).

4. object thrown moves at a height in meters per second is represented With h(t)=20t-4.9t^2: find the height in x seconds, if time increases by 2 how migher is the object, how high does it move per second, if time increased by h how much higher is it?
June 15th 2006, 11:23 PM
CaptainBlack

Quote:

Originally Posted by daniel4616

Hey, I have some question on college algebra
1. Is the line y=|x| symetric about the line y=x

No, it's symmetric about x=0.

For x>=0 it is the same as the line y=x
For x<0 it is the same as y=-x.

RonL
June 15th 2006, 11:26 PM
CaptainBlack

Quote:

Originally Posted by daniel4616

Hey, I have some question on college algebra

2. How do you prove points A(-1,2) B(3,-2) C(-6,-3) are vertices of a right triangle?

You compute the lengths of the sides AB, BC and AC, then show that
they satisfy Pythagoras' theorem (AB^2+AC^2=BC^2).

RonL
June 15th 2006, 11:34 PM
CaptainBlack

Quote:

Originally Posted by daniel4616

3. given f(x)= 2x if 0<x<1, 1-x if 1=<x<2 , 0 if 1=<x=<3: find the domain,f(1), f(2), f(3), and f(0.1).

A. You appear to have an error in the function definition, the way you have
defined it f(x) has two values for 1=<x<2. I will assume that the last
part of the definition should be: f(x)=0 2=<x=<3.

B. The domain is the set on which the function is defined. So in this case
f(x) is defined for all x such that 0<x=<3.

C. 1 lies in the second specified range so f(1)=1-1=0.
2 lies in the third specified range so f(2)=0, same for 3.
0.1 lies in the first specified range so f(0.1)=2x0.1=0.2.

RonL
June 15th 2006, 11:36 PM
CaptainBlack

Quote:

Originally Posted by daniel4616

4. object thrown moves at a height in meters per second is represented With h(t)=20t-4.9t^2: find the height in x seconds, if time increases by 2 how migher is the object, how high does it move per second, if time increased by h how much higher is it?

Try to reword this more closely to the original. As it is I would be guessing
at what you actually want to ask.

RonL
June 16th 2006, 04:46 AM
daniel4616

Quote:

Originally Posted by CaptainBlack

You compute the lengths of the sides AB, BC and AC, then show that
they satisfy Pythagoras' theorem (AB^2+AC^2=BC^2).

RonL

I know how to do it, the problem is that when I find the lengths, I dont get an euqation equal to a2+b2=c2
I get AB= sqrt32
BC= sqrt 82
AC=sqrt 50

The height of an object thrown in the air depends on the time it has been thrown. One situation the hieght in meters of an object after t seconds can be represented by h(t)=20t-4.09t^2. Whats the hieghts after x seconds, if time nicreases by 2 how much higher is the object, how much higher is the object per second nicrease? how much higher is the object per unit increase?
June 16th 2006, 05:26 AM
Soroban

Hello, daniel4616!

Quote:

1. Is the line $y=|x|$ symetric about the line $y=x$ ?

Did you make a sketch?

Code:

| * | * * | * * | * * | * - - - - * - - - - |

Quote:

2. How do you prove points A(-1,2) B(3,-2) C(-6,-3) are vertices of a right triangle?

An alternate method . . .

Find the slopes of AB, BC, and AC.

$m_{AB} = \frac{-2 - 2}{3-(-1)} = \frac{-4}{4} = -1$

$m_{BC} = \frac{-3-(-2)}{-6-3} = \frac{-1}{-9} = \frac{1}{9}$

$m_{AC} = \frac{-3 - 2}{-6-(-1)} = \frac{-5}{-5} = 1$

Since $m_{AB}$ and $m_{AC}$ are negative reciprocals: $AB \perp AC$

Therefore: $\angle BAC = 90^o$ and $\Delta ABC$ is a right triangle.

Quote:

#4 is very badly worded . . .
4. An object thrown is a height represented by: $h(t)\:=\:20t-4.9t^2$
(a) Find the height in $x$ seconds
(b) If time increases by 2 seconds, how much higher is the object? . . . should be asked last
(c) How high does it move per second? . . . How high? or "how far"?
(d) If time increased by $h$ , how much higher is it? . . . but h represents height

$(a)$ In $x$ seconds, the height is: $h(x)\:=\:20x - 4.9x^2$

$(d)$ Suppose time is increased from $t$ to $t + k$ .

Its height at time $t$ is: $h(t)\:=\;20t - 4.9t^2$ meters.

Its height at time $t+k$ is: $h(t+k)\;=\;20(t+k) - 4.9(t+k)^2$
. . . $= \;20t + 20k - 4.9t^2 - 9.8kt - 4.9k^2$

The difference is: $(20t + 20k - 4.9t^2 - 9.8kt - 4.9k^2) - (20t - 4.9t^2)$

Therefore, it is: $20k - 9.8kt - 4.9k^2$ meters higher.

$(b)$ If $k = 2$ , we have: $h(t + 2)\;=\;20(2) - 9.8(2)t - 4.9(2^2)\;=\;20.4 - 19.6t$ meters higher.
June 16th 2006, 06:55 AM
CaptainBlack

[QUOTE=daniel4616]I know how to do it, the problem is that when I find the lengths, I dont get an euqation equal to a2+b2=c2
I get AB= sqrt32
BC= sqrt 82
AC=sqrt 50
QUOTE]

AB^2+AC^2=32+50=82=BC^2

RonL