Easy
the easiest way is to imagine that you have a rectangle with a corner cut off
the area of the figure=area of rectangle-area cut off
area that was cut off was a triangle
area of rectangle=length times width
area of triangle=1/2base times height
so we draw imaginary lines like in attachment
find area of rectangle
legnth=9 inch
width=9 inch
area=9 times 9=81 in^2
triangle
the top part is ?+5=9, so the base of triangle=4
the side is ?+4=9, so the base is 5
area=1/2 times 4 times 5=10 in^2
area=rectangle-triangle
area=81-10=71 in^2
Since one equation has a negative y and the other has a positive y, I'm going to use those since they cancel each other out. Before that, the two y's need to be equal to each other.
x+2y=6
x-y=3
Multiply the bottom equation by two so then you have:
x+2y=6
2x-2y=6
The y's now cancel out:
x=6
2x=6
Add them together
3x=12
Divide
x=4.
To find y, plug x into either equation (*don't have to do both, but I will)
(4)+2y=6
(4)-y=3
Subtract four
2y=2
-y=-1
Divide each
2y/2 = 2/2
y=1
-y/-1 = -1/-1
y=1
The answer is:
x=4
y=1
I hope that helps!
Answer:
Q' (-3,4)
Step-by-step explanation:
A reflection across the x-axis would mean that only the x value of the point would change. This means that the y value would still be 4. To find the x value, just flip it to a negative. The negative of 3 is -3
I think it’s q12 or C hope this helped