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Artist 52 [7]
3 years ago
7

Kaitlyn uses conkainers shaped like a cylinder to catch rainwater. The dimensions of the containers are 12 ft in diameter and 9

ft in height. How much
rainwater can she collect if she has 6 containers?
Mathematics
1 answer:
Novay_Z [31]3 years ago
4 0

Answer:

1944 pi

or 6,104.6 (used 3.14 for pi)

Step-by-step explanation:

so the volume of one cylinder is

area of the base times hieght which is

36 pi *9=324 pi

so six containers

so 324 pi *6=1944pi

which is like 6,104.16

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6x - 2y = 26<br><br> 2x + 3y = 5<br><br> Solve for X and Y
Softa [21]

Answer:

y is equal to -1

x is equal to 4

Step-by-step explanation:

The first thing you would do is multiply the bottom equation by a negative 3. You will get - 6x - 9y = -15. Keep the top equation the same. Both of the 6x's cancel. You then get -11y = 11 so y is equal to negative 1.

Since you have Y, you can now plug that in to any equation to find the value of x. -6x - 2(-1) = 26. You're left with 4. Plug both of the values into each equation to double check.

7 0
3 years ago
Prove
Dennis_Churaev [7]

\frac{ \sin(a) -  \cos(a)  + 1 }{ \sin(a) +  \cos(a)  - 1 }  =  \\

____________________________________________

\frac{ \sin(a) -  \cos(a)  + 1 }{ \sin(a) +  \cos(a) - 1  }  \times  \frac{ \sin(a)  +  \cos(a)  + 1}{ \sin(a) +  \cos(a) + 1 }  =

\frac{ {sin}^{2}(a) + 2 \sin(a)  -  {cos}^{2} (a) + 1 }{ {sin}^{2}(a) + 2 \sin(a) \cos(a)  +  {cos}^{2}(a) - 1   }  =

_____________________________________________

As you know :

{sin}^{2} (a) +  {cos}^{2} (a) = 1

_____________________________________________

\frac{ {sin}^{2} (a) -  {cos}^{2}(a) + 2 \sin(a)   + 1}{ {sin}^{2} (a) +  {cos}^{2}(a) - 1 + 2 \sin(a)  \cos(a)  }  =

\frac{ {sin}^{2}(a) - (1 -  {sin}^{2}(a)) + 2 \sin(a) + 1   }{1 - 1 + 2 \sin(a)  \cos(a) }  =

\frac{ {sin}^{2} (a) +  {sin}^{2} (a) - 1 + 1 + 2 \sin(a) }{2 \sin(a) \cos(a)  }  =

\frac{2 {sin}^{2}(a) + 2 \sin(a)  }{2 \sin(a) \cos(a)  }  =

\frac{2 \sin(a)( \sin(a)  + 1) }{2 \sin(a)( \cos(a) \:  ) }  =  \\

\frac{ \sin(a)  + 1}{ \cos(a) }  \\

And we're done...

Take care ♡♡♡♡♡

6 0
3 years ago
Label each statement as true or false regarding the zeros/roots of a quadratic function. The roots zeros of a quadratic function
ikadub [295]

Answer:

True, false, true, true.

Step-by-step explanation:

The roots zeros of a quadratic function are the same as the factors of the quadratic function. This is true because your roots are your factors—>(x-3) is a factor, x=3 is the root.

The roots zeros are the spots where the quadratic function intersects with the y-axis. No! Those are called y-intercepts!

The roots zeros are the spots where the quadratic function intersects with the x-axis. True. X-intercepts are your solutions. (x-3) graphed would the (3,0). That’s a solution.

There are not always two roots/zeros of a quadratic function,​ True. No solution would be when your quadratic doesn’t intersect the x-axis. One solution would be when your vertex would be on the x-axis. Two solutions is when your quadratic intersects the x-axis twice. Can there be infinite solutions? No. It’s either 0, 1, or 2 solutions.

6 0
3 years ago
I need help is it A B C or D hopefully a pic uploaded
Maurinko [17]
The answer i came up with is A.
3 0
3 years ago
Evaluate the expression??<br> -5/k+1 for k =-1
vladimir2022 [97]

Answer:

undefined

Step-by-step explanation:

Order of Operations: BPEMDAS

Step 1: Define

-5/(k + 1)

k = -1

Step 2: Substitute and Evaluate

-5/(-1 + 1)

-5/0

undefined (we cannot divide anything by 0)

8 0
4 years ago
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