Answer:
Solving systems of equations with 3 variables is very similar to how we solve systems with two variables. When we had two variables we reduced the system down
to one with only one variable (by substitution or addition). With three variables
we will reduce the system down to one with two variables (usually by addition),
which we can then solve by either addition or substitution.
To reduce from three variables down to two it is very important to keep the work
organized. We will use addition with two equations to eliminate one variable.
This new equation we will call (A). Then we will use a different pair of equations
and use addition to eliminate the same variable. This second new equation we
will call (B). Once we have done this we will have two equations (A) and (B)
with the same two variables that we can solve using either method. This is shown
in the following examples.
Example 1.
3x +2y − z = − 1
− 2x − 2y +3z = 5 We will eliminate y using two different pairs of equations
5x +2y − z = 3
Step-by-step explanation:
Do you have a picture I can’t see it my guy
Create a conversion factor from "<span>There are 6.02 × 10^23 atoms in 1 gram of hydrogen:"
6.02*10^23 atoms
-------------------------
1 gram
Then mult. 800 grams by this conversion factor:
</span>6.02*10^23 atoms 800 grams
------------------------- * ----------------
1 gram 1
This comes out to 4816*10^23 atoms, which, in scientific notation, is
4.816*10^26 atoms.
Pretty sure the answer is 18
3^2+3^2=9+9=18 (with a square root)
Option C
The simplified form of given expression is 
<em><u>Solution:</u></em>
Given that we have to simplify the expression shown below
Let us first convert 0.00048 to scientific notation
<em><u>Steps to follow:</u></em>
Move the decimal point in your number until there is only one non-zero digit to the left of the decimal point. The resulting decimal number is a.
Count how many places you moved the decimal point. This number is b.
If you moved the decimal to the left b is positive.
If you moved the decimal to the right b is negative.
If you did not need to move the decimal b = 0.
Write your scientific notation number as a x 10^b and read it as "a times 10 to the power of b."
Remove trailing 0's only if they were originally to the left of the decimal point.
So the given expression becomes,
Let us make the exponent of second term as -4
Take 
as common term,
Thus option C is correct
