We have
f(x) = a(x – h)²<span> + k
we know the vertex v(5,3)
</span><span>substitute in the values for h and k
</span>f(x) = a(x – 5)²<span> + 3
</span><span>Use another point and substitute in values for x and f(x).
for the point (6,5)
</span><span>Solve for a.
5 = a(6 – 5)2 + 3-------------- > 5=a+3-------------> a=2
</span>
The function is f(x)=a(x – h)2 + k-------- > 2(x – 5)² + 3
f(x)= 2(x – 5)² + 3-------- > 2[x²-10x+25]+3=2x²-20x+50+3=2x²-20x+53
f(x)=2x²-20x+53
<span>
the answer is f(x)=</span> 2(x – 5)² + 3----------------- > (f(x)=2x²-20x+53)<span>
</span>
You multiply the two integers together
Area of Rhombus =
Diagonal DB = 2×8 = 16
To find AC, we substitute into the formula the area 168 in² and DB of 16 in
Area =
inches
Answer: B
Answer:
I wish I could help but I wasn't very good at geometry
