Ask question

Ask question

All categories

3 years ago

7

PLEASE HELP DUE TODAYYYY

1 answer:

s2008m [1.1K]3 years ago

8 0

For the second one it should be 12 or -12

You might be interested in

Pls help being timed

Alex777 [14]

The answer is B your welcome

5 0

3 years ago

The Spirit Team will be selling popcorn as a fundraiser at the next basketball game to earn funds to go to Orlando for a competi

Bogdan [553]

Answer:1 and 2

Step-by-step explanation:

Given

Volume needed $V=220\ in.^3$

For first cylinder

$r=3\ in.$

$h=4\ in.$

$V=\pi r^2h$

$V_1=\pi (3)^2(4)$

$V_1=113.11\ in^3$

For second cylinder

$r=4\ in.$

$h=4\ in.$

$V=\pi r^2h$

$V_2=\pi (4)^2(4)$

$V_2=201.08\ in^3$

For Third cylinder

$r=9\ in.$

$h=9\ in.$

$V=\pi r^2h$

$V_3=\pi (9)^2(9)$

$V_3=2290.51\ in^3$

For fourth cylinder

$r=4\ in.$

$h=9\ in.$

$V=\pi r^2h$

$V_4=\pi (4)^2(9)$

$V_4=452.448\ in^3$

So, cylinder $1$ and $2$ are appropriate as it is under $220\ in.^3$

3 0

3 years ago

Find the Derivative y’ implicitly.

Brums [2.3K]

$e^{x^2y} - e^y = e^y(e^{x^2} - 1) = x$

<em>We should ISOLATE x</em>

<em /> $e^y= \frac{x}{e^{x^2} - 1}$

<em>Find the Natural Log of Both Sides to Make the Left Side "y"</em>

<em /> $y = ln(\frac{x}{e^{x^2}-1})$

<em>Now, FIND THE DERIVATIVE Using Chain Rule!!!</em>

<em /> $y' = \frac{1}{\frac{x}{e^{x^2}-1}} * \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2}= {\frac{e^{x^2}-1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2} = {\frac{1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)} = {\frac{1}{x}}* \frac{(e^{x^2}-1 - 2x^2e^{x^2})}{(e^{x^2}-1)}$

3 0

3 years ago

What is the approximate volume of the cylinder?

Delicious77 [7]

Answer:

V ≈ 509 m3 ← with 3.14 for π =2462 feet³

Step-by-step explanation:

I hope I helped, This is my math just round up :)

3 0

3 years ago

If f(x)=2x-5, find <br><br> a. f(4)<br> b. f(-3)

Vladimir [108]

Replace x in the given equation with each value:

a. f(4) = 2(4) -5

= 8-5

=3

b. (f-3) = 2(-3) - 5

= -6 -5

= -11

8 0

3 years ago