Answer:
$141.75
Step-by-step explanation:
189*.75
1-.25 b/c 25 percent off
I believe you should try to plug in the ordered pair to see if the solutions work.
The ordered pair is (x,y) because x is on the x-axis and y is on the y-axis, therefore in (16,-3) the x=16 and y=-3.
Now substitute, meaning plug in those numbers into the equation.
For x+2y=10 would be 16+2(-3)=10
Now you just need to solve one side and see if it equals the other side.
16+2(-3)
First you use order of operations to solve this. PEMDAS. So you multiply 2(-3) first because of the parenthesis and it being multiplication.
16+(-6) and when you have a positive number adding a negative number it’s going backwards of the number line, basically meaning subtraction in a way. Sorry if this confuses you, if you already know how to do negatives and such nevermind this part.
But 16+(-6)=10
So now you look at both side of the equation, does the left side equal to the right? 10=10, so yes. It is a solution for that equation.
Now for the next equation, 7y=-21
Again, plug in the ordered pair (16,-3) into the equation. Remember that it’s (x,y).
There is no x in this equation so no need to worry about that; you only plug in y for this one.
7(-3) Now you multiply. Whenever you multiply a positive number and a negative number, the answer will always be negative. So 7(-3) is -21.
Now look if the left side is equivalent to the right. Does -21=-21? Yes. The ordered pair is a solution.
(16,-3) is a solution to both equations.
Hope this helps!
Answer:
The coefficient is -5, and the base is 12.
Step-by-step explanation:
The coefficient is the number that the variable is being multiplied by in an expression, so -5 would be the coefficient in the expression 
. The base would be the little subscript number that is next to the variable, so the base in the expression is 12.
Answer:
x+3
Step-by-step explanation:
sqrt(x^2+6x+9)
factor
We know this is a perfect square trinomial
sqrt(( x+3)^2)
Taking the square root of a square
x+3
I forget how to do this. I haven't done this in awhile.
