The standard form of the equation of a parabola is x = y2 + 6y + 1. What is the vertex form of the equation? O A. x = (y + 3)2 - 8 O B. x = (y + 6)2-11 O C. x = (y + 6)2-35 O D. x = (y + 3)2-5​

A

Step-by-step explanation:

Given

x = y² + 6y + 1

To complete the square

add/subtract ( half the coefficient of the y- term )² to y² + 6y

x = y² + 2(3)y + 9 - 9 + 1

  = (y + 3)² - 8

sec ф = 5/3 (quadrant 4)

let x = base

y = perp

z = hypotenuse

sec ф = hypotenuse/base = 5/3

⇒ z = 5 and x = 3

finding y:

z² = x² + y²

5² = 3² + y²

y² = 25-9

y² = 16

y = 4

now finding all other geometric functions of theta:

sin ф = y/z = 4/5

sin ф = -4/5 (sin is negative in the 4th quadrant)

cos ф = x/z = 3/5

cos ф = 3/5 (cos is positive in 4th quadrant)

tan ф = y/x = 4/3

tan ф = -4/3 (tan is negative in 4th quadrant)

cosec ф = z/y = 5/4

cosec ф = -5/4 (cosec is negative in the 4th quadrant)

cot ф = x/y = 3/4

tan ф = -3/4 (tan is negative in 4th quadrant)

12

step-by-step explanation:

a triangle's   full angle is 180 degrees so 180 - 156 = 24. 24/2=12