The vertex of this parabola is at (5, 5). when the y-value is 6, the x-value is -1. what is the coefficient of the squared term in the parabola's equation?

1/36  

Step-by-step explanation:

The vertex form of the equation for a parabola is

y = a(x - h)² + k

where h and k are the coordinates of the vertex

and a is the coefficient of x² in the standard form.

Data:

Vertex at (5,5)

A point at (-1,6)

Calculations:

1. Substitute the coordinates of the vertex into the equation

y = a(x - 5)² + 5

2. Substitute the coordinates of the point into the equation and solve for a

6 = a(-1 -5)² + 5

6 = a(-6)² + 5

6 = 36a + 5

1 = 36a

a = 1/36

The coefficient of x² in the standard form is 1/36.

step-by-step explanation:

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92