The figure shown is a rhombus. What condition would also make it a square? A. The figure would be a square if the diagonals were
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Answer:
A) The vertex ( h , k) = ( -2 , -18)
B) The minimum value of the given function = - 18
C) The Axis of the symmetry for f(t) is y -axis
Step-by-step explanation:
A)
Given a parabola f(t) = t² + 4 t − 14
f(t) = t² + 2(2) t + (2)²-4− 14
f(t) = (t +2)² - 18
Let comparing y = (x +2)² -18
(x +2)² = y + 18
(x-h))² = 4 a ( y - k))²
<em>The vertex ( h , k) = ( -2 , -18)</em>
B)
Given a parabola f(t) = t² + 4 t − 14
Differentiating with respective to 't'
f¹(t) = 2 t + 4
f¹(t) = 2 t + 4 = 0
now
Again Differentiating with respective to 't'
f(x) has a minimum value at t = -2
Given f(t) = t² + 4 t − 14
f( -2) = 4 + 4(-2) -14 = 4 -8 -14 = -18
The minimum value of the given function = - 18
C)
f(t) = (t +2)² - 18
Let comparing y = (x +2)² -18
(x +2)² = y + 18
(x-h))² = 4 a ( y - k))²
The vertex ( h , k) = ( -2 , -18)
The Axis of the symmetry for f(t) is y -axis
well, Matti's house is a triangular prism, and to get the volume of it, we simply get the area of the triangle upfront and multiply by its length of 15.