Equations with absolute value:

Where k is a positive number; if k is a negative number, the equation is impossible (absolute value is always positive).
How to solve:


Then:
1. |x+7|=12
x+7=12 V -x-7=12
x=5 V -x=19
x=5 V x=-19
{-19, 5}
2. |2x+4|=8
2x+4=8 V -2x-4=8
2x=4 V -2x=12
x=2 V x=-6
3. 3|3k|=27
3×3k=27 V 3×(-3k)=27
9k=27 V -9k=27
k=3 V k=-3
{-3, 3}
4. 5|b+8|=30
5×(b+8)=30 V 5×(-b-8)=30
5b+40=30 V -5b-40=30
5b=-10 V -5b=70
b=-2 V b=-14
{-14, -2}
5. |m+9|=5
m+9=5 V -m-9=5
m+9=5 V m+9=-5
The answer is table C
For anything to be a function, it cannot have two pairs as my teacher taught me!
For example, let’s take the number 4
Let’s say 4 can take a friend to the fair!
4 decided to take 5 (4,5)
4 CANNOT take anyone else but 5
4 is allowed to take as many 5’s as they want but cannot take any other number
Let’s set up a table
X. Y
4. 5
4. 5
4. 5
This is a function
If the table changed:
X. Y.
4. 5
4. 5
4. 6
This will not be a function
I hope this helps!
219. If our rounding down
...........
Put it into standard form so you would have
27x10¯6 divided by 9x10¯6
Then you can divide the numbers so 27/9 = 3
Then to divide the powers you just take them away so - 6--6 = - 6 +6 = 0
Therefore the answer is just 3
214.3125
Happy to help (Mark brainliest)