Vertex at the origin and opening down → y=ax^2
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4
y=ax^2→y=-(1/4)x^2
7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4
Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m
Please, see the attached file.
Answer: Third option 15.75 m
Answer:
<h3>
f(x) = - 3(x + 8)² + 2</h3>
Step-by-step explanation:
f(x) = a(x - h)² + k - the vertex form of the quadratic function with vertex (h, k)
the<u> axis of symmetry</u> at<u> x = -8</u> means h = -8
the <u>maximum height of 2</u> means k = 2
So:
f(x) = a(x - (-8))² + 2
f(x) = a(x + 8)² + 2 - the vertex form of the quadratic function with vertex (-8, 2)
The parabola passing through the point (-7, -1) means that if x = -7 then f(x) = -1
so:
-1 = a(-7 + 8)² + 2
-1 -2 = a(1)² + 2 -2
-3 = a
Threfore:
The vertex form of the parabola which has an axis of symmetry at x = -8, a maximum height of 2, and passes through the point (-7, -1) is:
<u>f(x) = -3(x + 8)² + 2</u>
5.25 / .85 you would get 6.17~ but you cant sell 1/4th a cup, so you have to round up, making it 7 cups.
Answer:
I think the ans is option 'a'
Final Answer:
Corresponding Angles Theorum; ∠AGF and ∠EHD are congruent
Answer:
I cant see the picture
Step-by-step explanation: