Select all the intervals where f(x)= -x^3 + 3x^2 +1 is only decreasing.

-infinity < x < 1
5 < x < infinity
-infinity < x < 0
-infinity < x < infinity
2 < x < infinity
0 < x < 2
1 < x < 5

5 < x < ∞

-∞ < x < 0

2 < x < ∞

Step-by-step explanation:

Given function is + 3 + 1.

To, Calculate where the function is increasing or decreasing,

we have to calculate the derivative of the function,

so,

f'(x) = -3 + 6x.

Equating the f'(x)= 0 , we get

x(-3x + 6) = 0

So, f'(x) will be zero at x = 0 and x = 2.

f(x) will be decreasing in the interval where f'(x) will be negative.

now as the coefficient of f'(x) is negative it will be negative between

the interval -∞ < x < 0 and 2 < x < ∞.

Options which lie in these intervals only are

5 < x < ∞

-∞ < x < 0

2 < x < ∞

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)

b. the decimal could be two-tenths.

step-by-step explanation: