This is the new update of the question

d

Step-by-step explanation:

If x represents the slant height, the graph would be a straight line.

If x represents the radius, the graph would be a parabola.

Looking at the answers, we're looking for x to be the slant height, so we're looking for a straight line.

We are looking for g(x). Answers a and b are wrong.

In g(x) r is fixed to 5 (this is given). So the slant height in the smallest possible cone would still be at least 5.

So the domain is x > 5.

d is your answer.

We have this formula for the surface area of a cone

SA = pi*r*s + pi*r^2

The pi*r^2 portion is the flat bottom part of the cone, which is a circle of radius r. The pi*r*s portion is the lateral surface area. This is the triangular part of the cone, so to speak.

"one of the functions represents the surface area of a cone with radius 5 and slant height x". So we can say r = 5 and s = x to go from this

SA = pi*r*s + pi*r^2

to this

SA = pi*5*x + pi*5^2

SA = 5pi*x + 25pi

y = 5pi*x + 25pi

note how this is a linear function in the form y = mx+b

"The other function represents the surface area of a cone with slant height 5 and radius x", meaning s = 5 and r = x

SA = pi*r*s + pi*r^2

SA = pi*x*5 + pi*x^2

SA = 5pi*x + pi*x^2

SA = pi*x^2 + 5pi*x

y = pi*x^2 + 5pi*x

This is a quadratic function due to the x^2 term. This graphs out a parabola.

The graph given to us has f(x) as the parabola (it's the bowl shaped graph) while g(x) is the straight line

Therefore,

f(x) = pi*x^2 + 5pi*x

g(x) = 5pi*x + 25pi

f(x) is the cone with radius x and slant height 5

g(x) is the cone with radius 5 and slant height x

Keep in mind that the slant height of a cone forms the hypotenuse of a right triangle when you do a vertical cross section. If the radius is 5, then the leg of the triangle is 5. The hypotenuse is always longer than either leg, so the hypotenuse x must be larger than 5. A right triangle would not be possible with the hypotenuse x < 5 (you would have to shrink the radius).

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)

sorry, can't i suggest going to algebra.com

step-by-step explanation: