Plz help!!! Will give brainliest!!!

cricket20 [7]

There’s a pattern to this and I’ll explain in the comments if you want to know or can’t figure it out :)

1.) 27/x^3

2.) a^8/b^12

3.) 7^36/81 OR 13,841,287,201/81

4.) 256m^20/n^4

5.) x^6y^6

6.) 64k^4/k^6

7.) 1,000a^9b^6/a^6b^9

8.) 81x^21/625

9.) 2304/x^16y^12

10.) x^7z^14/y^21

11.) 7,776k^5

12.) x^24y^24/4,096

Hope it’s all helped

7 0

2 years ago

What is the answer for this problem

solong [7]

I don't know what your story is, but in a story, usually a problem would build tension.

Hope this helps!! :)

6 0

3 years ago

Find the volume of a cylinder that has a diameter of 6 and a height of 9. Leave your answer in terms of (pie)

laiz [17]

V = π x r² x h

V= π x 6² x9

V= 324 π

5 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) = $0.78^{4}ln(0.78)$