V-5= 3(x+1) in slope intercept Form
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I think that you should imagine what regular shape the figure would have been.
The fact that two sides are 9inches makes me think it would have been a square.
The area of a square is A= l² where l is the length of one side and A=area
A=9*9
A=81 in.²
Next, find the area of the part that is taken away,(it looks like a triangle). to find that, if the figure was a square, the length on top would be 9 not 5, so the length gone is 9-5=4, so the base of the triangle is 4,
If the figure was a square, the length on the other side would be 9 not 4 so the length gone is 9-4=5, so the height of the triangle is 5
the area of a triangle is A= 1/2bh, where A is area, b is base and h is height.
A=1/2*4*5 A=10 so the area of the part gone is 10 inches²
The area of the figure is the area of the square -the area of the triangle
A=81-10
A=71inches²
,
The fact that two sides are 9inches makes me think it would have been a square.
The area of a square is A= l² where l is the length of one side and A=area
A=9*9
A=81 in.²
Next, find the area of the part that is taken away,(it looks like a triangle). to find that, if the figure was a square, the length on top would be 9 not 5, so the length gone is 9-5=4, so the base of the triangle is 4,
If the figure was a square, the length on the other side would be 9 not 4 so the length gone is 9-4=5, so the height of the triangle is 5
the area of a triangle is A= 1/2bh, where A is area, b is base and h is height.
A=1/2*4*5 A=10 so the area of the part gone is 10 inches²
The area of the figure is the area of the square -the area of the triangle
A=81-10
A=71inches²
,
times both sides by 3
x+5≤6
minus 5 both sides
x≤1