Which of these trends represents an inverse variation?

Please help, il mark brainliest for the right answer

kari74 [83]

Answer:

I think that it may be 33

Because, you see 15-9=6

so you do 20-9=11

Then you do 25-9=16

add them all together to get 33 so i think that's the perimeter

I am really not sure if this is correct...i'm kinda bad at stuff like this

6 0

3 years ago

evelyn rode her horse along a triangular path. The distance she traveled south was 5 times the distance she traveled east. Then

cricket20 [7]

(5x)^2 + x^2 = c^2

25x^2 + x^2 =c^2

sqrt 26 x ^ 2 = sqrt c^2

x sqrt 26= c

so answer is 6x + x sqrt 26

8 0

4 years ago

Recently in the United States, there were about 300,000,000 cell phone users. That same year there were about 5.7x10^9 cell phon

Alla [95]

$\frac{users worldwide}{US users}$
$\frac{5700000000}{300000000}$
$19$

6 0

3 years ago

Read 2 more answers

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre

Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = $\frac{d}{dx}$ [ $x^{4}ln(x)$ ]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = $4x^{3}ln(x) + x_{4}.\frac{1}{x}$

f'(x) = $4x^{3}ln(x) + x^{3}$

f'(x) = $x^{3}[4ln(x) + 1]$

Now, find the critical points: f'(x) = 0

$x^{3}[4ln(x) + 1]$ = 0

$x^{3} = 0$

x = 0

and

$4ln(x) + 1 = 0$

$ln(x) = \frac{-1}{4}$

x = $e^{\frac{-1}{4} }$

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval x-value f'(x) result

0<x<0.78 0.5 f'(0.5) = -0.22 decreasing

x>0.78 1 f'(1) = 1 increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = $x^{4}ln(x)$

f(0.78) = $0.78^{4}ln(0.78)$