# Math Help - College Algebra problems..

1. ## 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?

2. Originally Posted by daniel4616
Hey, I have some question on college algebra
1. Is the line y=|x| symetric about the line y=x

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

RonL

3. 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

4. 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

5. 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

6. 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?

7. Hello, daniel4616!

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

Did you make a sketch?
Code:
              |
*       |       *
*     |     *
*   |   *
* | *
- - - - * - - - -
|

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.

#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.

8. [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