Easy
y=a(x-h)^2+k
vertex is (h,k)
we know that vertex is (4,0)
input that point for (h,k)
y=a(x-4)^2+0
y=a(x-4)^2
passes thorugh the point (6,1)
input that point to find a
1=a(6-4)^2
1=a(2)^2
1=a(4)
divide both sides by 4
1/4=a
thefor the equation is
y=(1/4)(x-4)^2
or
y=(1/4)x^2-2x+4
Step-by-step explanation:
I'll do the first problem as an example.
∠P and ∠H both have one mark. That means they're congruent.
∠T and ∠G both have two marks. So they're congruent.
∠W and ∠D both have three marks. So they're congruent.
So we can write a congruence statement:
ΔPTW ≅ ΔHGD
We can write more congruence statements by rearranging the letter, provided that corresponding pairs have the same position (P is in the same place as H, etc.). For example:
ΔWPT ≅ ΔDHG
ΔTWP ≅ ΔGDH
Rounded to the nearest tenth, the square root of 149 is 12.2.
The correct answer would be 2y^5
Answer:
n = 0
Step-by-step explanation:
1.2n + 1 = 1 - n
Add n and - 1 on both sides.
1.2n + n = 1 - 1
Combine like terms.
2.2n = 0
Divide both sides by 2.2.
n = 0