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I need help please (brainllest)
1 answer:
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Answer:
Step-by-step explanation:
X^2-6x+5=x^2-5x-x+5=x(x-5)-(x-5)=(x-5)(x-1)
Then we have (x-5)(x-1)=0 if x=5 or x=1.
Intersections on x are points (1,0) and (5,0), middle is (3,0).
Intersection with y is when put x=0 in equation, so you will get y=0-6*0+5, y=5. The point is (0,5).
From picture symmetry is line x=3.
A square has 4 sides of equa length
the perimiter is 4 times the side, or P=4s where s=length of 1 side
the area is length times witdth, but since the length=width, area=s² where s is the length of one side
so express A=s² in terms of P when P=4s
solve for s first in P=4s
P=4s
divide both sides by 4
subsitute
for s in the other eqaution
in terms of the perimiter, the area is
the perimiter is 4 times the side, or P=4s where s=length of 1 side
the area is length times witdth, but since the length=width, area=s² where s is the length of one side
so express A=s² in terms of P when P=4s
solve for s first in P=4s
P=4s
divide both sides by 4
subsitute
in terms of the perimiter, the area is