If CD is not parallel to EF, what is m2C?

Im having trouble. can someone please help me with this !

VladimirAG [237]

Answer:

28

Step-by-step explanation:

5 0

3 years ago

Please help it’s for a unit test!

notka56 [123]

Answer:

X=___ or x=8 or one of my other answers in the explanation! I hope this helped!

Step-by-step explanation:

it must be 4 because 4 goes on the side by +4 and its x+4 then the other side of the shape is 8 so that meaning x is 4 then 2x+1 is 2 times X + 1 so 2 times 4... 2,4,6,8 or 4+4=8 so then 2 times + 1 is 8 + 1 so 8+1=9 so 2x+1 might also be 12 because its on the other side of the problem/ shape so you use all that you have so you could maybe plus the other side not the problem side for making it a and c and stuff but also 8+12 so 8+12=20 then x might be 20 but you answer meaning X is 8. X=8 Brainliest? i tried really hard! I am not even near this grade whatever grade this is... Plz! Thanks! Thank you! Have a great day or week or weekend! i answer a lot of questions and help 114 people or something or more! Plz put 5 stars, a heart, and brainliest i would really appreciate that thank you! Enjoy the rest or your day! Bye! <3 :)

7 0

3 years ago

A metal cylinder can with an open top and closed bottom is to have volume 4 cubic feet. Approximate the dimensions that require

Aleksandr-060686 [28]

Answer:

$r\approx 1.084\ feet$

$h\approx 1.084\ feet$

$\displaystyle A=11.07\ ft^2$

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 $ft^3$ . The volume of a cylinder is given by

$\displaystyle V=\pi r^2h$

Equating it to 4

$\displaystyle \pi r^2h=4$

Let's solve for h

$\displaystyle h=\frac{4}{\pi r^2}$

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

$\displaystyle A=\pi r^2+2\pi rh$

Replacing the formula of h

$\displaystyle A=\pi r^2+2\pi r \left (\frac{4}{\pi r^2}\right )$

Simplifying

$\displaystyle A=\pi r^2+\frac{8}{r}$

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

$\displaystyle A'=2\pi r-\frac{8}{r^2}=0$

Rearranging

$\displaystyle 2\pi r=\frac{8}{r^2}$

Solving for r

$\displaystyle r^3=\frac{4}{\pi }$

$\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet$

Computing h

$\displaystyle h=\frac{4}{\pi \ r^2}\approx 1.084\ feet$

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

$\displaystyle A''=2\pi+\frac{16}{r^3}$