1. Find the number of digits in the expansion of (2^120)(5^125) without any calculator or computer.

2. Find the coordinates of the two points that trisect the line segment with the endpoints (A(2,3) B(8,2).

l l = Absolute value
3. Give the function f(x) = x^3 - 2x, sketch y = f(lxl).
Sketch g(x) = lx^2 - 1l - lx^2 - 4l
Sketch the region in the plane to show all points (x,y)
such that lxl + lyl <= 2

2. Hello, shane99!

1. Find the number of digits in the expansion of: $(2^{120})(5^{125})$ without any calculator or computer.

We have: . $2^{120}\cdot 5^{125} \;=\;2^{120}\cdot5^{120}\cdot5^5 \;=\;(2^{120}\cdot5^{120})\cdot5^5 \;=\;(2\cdot5)^{120}\cdot5^5 \;=\;10^{120}\cdot5^5$

This is: . $5^5 = 3125$ followed by 120 zeros.

Therefore, the product has 124 digits.

3. Originally Posted by shane99
2. Find the coordinates of the two points that trisect the line segment with the endpoints A(2,3) and B(8,2).

l l = Absolute value
3. Give the function f(x) = x^3 - 2x, sketch y = f(lxl).
Sketch g(x) = lx^2 - 1l - lx^2 - 4l
Sketch the region in the plane to show all points (x,y)
such that lxl + lyl <= 2
2. Here's one way I can think of, but someone else could probably find a simpler way.
Find the equation of the straight line joining A and B.
You want to trisect the line AB, so find 1/3 of the distance between A and B, call it r.
Find the equations of the circles centred at A and B with radius r.
Find where the circles intersect the straight line joining A and B. Take the coordinates of the relevant points (the straight line will intersect the two circles at four points, but you're only looking for two points).

3. f(lxl) = |x|^3 - 2|x| = |x^3| - 2|x|
Draw x^3 and then draw |x^3|, that is, where x^3 is negative you 'flip' it about the x-axis (y=0) so it becomes positive.
Draw 2|x|.
Subtract the two graphs (you have to have drawn the two graphs on the same set of axes).

Try sketching g(x) yourself but doing similar reasoning as above.

For $|x|+|y|\leq 2$ use the definition of the absolute value and look at the different cases, i.e.
x<0 and y<0
x<0 and y>0
x>0 and y<0
x>0 and y>0
For instance, in the region x<0 and y>0 we have |x|=-x and |y|=y, and so $|x|+|y|\leq 2$ becomes $-x+y\leq 2$ and sketch this only in the region x<0 and y>0.

4. 1. Find the number of digits in the expansion of (2^120)(5^125) without any calculator or computer.
If I may give my 2 cents. A way to find the number of digits in a huge number is use logs.

We have $log(2^{120}\cdot 5^{125})=log(2^{120})+log(5^{125})=120log(2)+125lo g(5)\approx 123.494850022$

Round up and we see there are 124 digits.

To find the first few digits of this number. Take $10^{\text{the decimal part of the above}}$

We get $10^{.494850022}=3.12500000$

As Soroban said, it is 3125 and then 120 0's.

Though you may need a calculator, it is a cool way to know and I thought I would show it anyway.

5. Originally Posted by Soroban

This is: .[tex]5^5 = 3125

How do you know that 5^5= 3125 without using a calculator?

6. 5*5=25, 25*5=125, 125*5=625, 625*5=3125.

Do it the old-fashioned way sans calculator.

7. Originally Posted by shane99
3. Give the function f(x) = x^3 - 2x, sketch y = f(lxl).
Sketch g(x) = lx^2 - 1l - lx^2 - 4l
Sketch the region in the plane to show all points (x,y)
such that lxl + lyl <= 2
Alternatively,
f(|x|) = |x^3| - 2|x|
use the definition of the absolute value and look at the different regions:
|x^3| = -x^3 when -x^3<0, i.e. when x<0
|x^3| = x^3 when x^3>0, i.e. when x>0

|x| = -x when x<0
|x| = x when x>0

So for x<0
f(|x|) = -x^3 +2x
and for x>0
f(|x|) = x^3-2x

g(x) is done similarly, but is trickier because the regions you have to consider will be more complicated.

8. Need more help with 3c)