I can't exactly SHOW you where to put the numbers, but I can teach you the process of how you'd do it.
First off, label your number line from 0-15, as it is the simplest. (You'd be counting by 1 per each line). Then, follow this process:
1) Look at the first digit of your value. Place your number according to your first digit. (So, you'd put 0.365 at the 0 line and 3.521 at the 3 line)
2) Look at the second digit of your value. Imagine that between the two main lines (0-1 and 3-4) that there is 10 smaller lines. Then, you can place your number according to your second digit. (So, you'd put 0.365 at the 0.3 line and 3.521 at the 3.5 line).
3) Look at the third digit of your value. Imagine that between the two smaller lines (0.3-0.4 and 3.5-3.6) that there is 10 smaller lines. Then, you can place your number according to your third digit. (So, you'd put 0.365 at the 0.36 line and 3.521 at the 3.52 line).
4) Look at the fourth digit of your value. Imagine that between the two even smaller lines (0.36-0.37 and 3.52-3.53) that there is 10 smaller lines. Then, you can place your number according to your fourth digit. (So you'd place 0.365 at the 0.365 line and 3.521 at the 3.521 line)
Answer:
y = (∛x)/3
Step-by-step explanation:
To undo the multiplication by 27, you multiply by its inverse:
(1/27)y = (1/27)(27x^3)
y/27 = x^3 . . . . . . . . . . . simplify
To undo the cube, you take the cube root:
(∛y)/(∛27) = ∛(x^3)
(∛y)/3 = x
Apparently, you want the inverse function, so you swap the variables:
y = (∛x)/3
_____
You can swap the variables at the beginning or end. It doesn't matter. If you do it at the beginning, you have ...
x = 27y^3
and you're solving for y. You use the same inverse operations that we used above.
4x+24=84 4x=60 x=15 length =27 width = 15
Answer:
given
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>Optimizing With Derivatives
</u>
The procedure to optimize a function (find its maximum or minimum) consists in
:
- Produce a function which depends on only one variable
- Compute the first derivative and set it equal to 0
- Find the values for the variable, called critical points
- Compute the second derivative
- Evaluate the second derivative in the critical points. If it results positive, the critical point is a minimum, if it's negative, the critical point is a maximum
We know a cylinder has a volume of 4 
. The volume of a cylinder is given by
Equating it to 4
Let's solve for h
A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is
Replacing the formula of h
Simplifying
We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero
Rearranging
Solving for r
![\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet](https://tex.z-dn.net/?f=%5Cdisplaystyle%20r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4%7D%7B%5Cpi%20%7D%7D%5Capprox%201.084%5C%20feet)
Computing h
We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative
We can see it will be always positive regardless of the value of r (assumed positive too), so the critical point is a minimum.
The minimum area is
