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Molodets [167]
3 years ago
10

A cylindrical water bottle has a diameter of 5 in. and a height of 9 in. What is the capacity of the water bottle in cubic inche

s? Use 3.14 to approximate Pi and round your answer to the nearest tenth.
Mathematics
1 answer:
VikaD [51]3 years ago
3 0
The answer will be 388.6 because you will use the formula V=pi.r^2.h.
pi we know is 3.14 we also know that radius is half the diameter so half of 5 is 2.5 and the height is 9 so here is what we have so far. V=3.14x2.5^2x9.
2.5x2.5=13.75 13.75x3.14=43.175 43.175x9=388.575 and if we round that it will be 388.6 so your answer is 388.6
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Find the number of possible 5 card hands that contain the cards specified the cards are taken from a standard 52 card deck
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Answer:

Step-by-step explanation:

Find the number of possible 5-card hands that contain the cards specified. The cards are taken from a standard 52-card deck.

5 red cards.

Ans: 26C5 = 26!/[21!*5!] = 65780

--------------------------------------------

4 spades and 1 card that is not a spade.

Ans: [13C4*39C1] = 27885

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3 face cards (kings, queens, or jacks) and 2 cards that are not face cards.

Ans: 12C3*40C2 = 171,600

============================

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Please help me with #1. I need an equation and the number of years.
weeeeeb [17]
1. 12000(1 - n/5) = 2,400
     12000 - 2400n = 2400
      - 2400n + 12000 - 12000 = 2400 - 12000
      - 2400 = - 9600
             n = 4, After 4 years the scanner will be replaced.



2.  Let x represents Michael age in years now.
     
                            
                          Now           4years later  
                            
Michael                 x                 x + 4

      
John                      x - 17           (x - 17) + 4


                              x + 4 + x - 17 + 4 = 49
                               x + 4 + x - 13 = 49
                                2x - 9 = 49
                                 2x - 9 + 9 = 49 + 9
                                            2x = 58
                                     (1/2)(2x) = (1/2)(58)
                                                x = 29, Michaels age is 29


3. Brady speed 1   Rate(mph)       Time(hours)           Distance(miles)
                                17                     t                               17 t
     Brady speed 2    22                     4 - t                         22(4 - t)total distance

The total distance he rode  17t + 22(4 - t) = 70

The total distance = 70 miles


                                         showing work:  
                                           17t + 22(4 - t) = 70
                                            17t + 88 - 22t = 70
                                             - 5t + 88 = 70
                                              - 5t + 88 - 88 = 70 - 88
                                               - 5t  = - 18
                                                t = 3.6, Brady rode 17(mph) for 3.6(hrs).


4. t = the price of 1 shirt
     4t - 39.58 = 52.22
     4t - 39.58 = 52.22 + 39.58
      4t  + 0 = 91.80
    (1/4)(4t) = (1/4)(91.80)
               t = 22.95, price of 1 shirt was $22.95


5. Let M represent number of months.
      3.5m + 90 = 132
       3.5m + 90 - 90 = 132 - 90
        3.5m = 42
    (1/3.5)(3.5m) = (1/3.5)(42)
                      m = 12, the boy will grow 132cm tall, 12 months from now



6.   Let N represent given number
         n + 1/6n + (2 1/2)n + 7 = 12 1/2
          n ( 1 + 1/6 + 5/2) + 7  = 12 1/2
          n(6/6 + 1/6 + 15/6) + 7 = 12 1/2
           n(22/6 + 7 - 7 = 12 1/2
            11/3n + 7 - 7 = 12 1/2 - 7
              11/3n + 0 = 5 1/2
              11/3n = 5 1/2
               3/11.11/3 = 3/11.11/2
                 1n  = 3/2
                    n = 1 1/2, is the number


7. let X represents the first number, then 33 - x represents the other # since their sum is 33
     X - (33 - x) = 2
      X + ( - (33 - x)) = 2
      X + ( -33) + X = 2
       2x + (- 33) = 2
        2x + (- 33) + 33 = 2 + 33
        2x + 0 = 35
        1/2 * 2x = 1/2 * 35
         1x = 35/2
          x = 17 1/2

33 - x = 33 - (17 1/2) = 15 1/2
                     {17 1/2,  15 1/2}
                                 



8. Let T represents number of tokens in machine A. Then 1/2T represents the number of tokens in machine B, and 3/4T represents the number of tokens in machine C.

   T + 1/2T + 3/4T = 18,324
    9/4T = 18, 324
          T = 8, 144
Machine A took 8,144T, Machine B 4,072T, Machine C 6,108T



9.  Let Z represents the cost of a pencil in dollars. Then, the cost of a pen in dollars is Z + 0.11

   250(Z + Z + 0.11) = 42.5
    250(2Z + 0.11) = 42.5
    500Z + 27.5 = 42.5
    500Z + 27.5 + ( - 27.5) = 42.5 + ( -27.5)
     500Z + 0 = 15
     500Z = 15
      500Z/500 + 15/500
             Z = $0.03, a pencil costs, and  a pens cost $0.14


10. Let R represents speed in mph of faster car, Then r - 7 represents speed in mph of slow car.

 5 1/2(r) + 5 1/2(r - 7) = 599.5
  5 1/2(r + r - 7) = 599.5
 5 1/2(2r - 7) = 599.5
 11/2 (2r - 7) = 599.5
  2/11 * 11/2 (2r - 7) = 2/11 * 599.5
  1 * (2r - 7) = 1199/11
   2r - 7 = 109
   2r  - 7 + 7 = 109 + 7
   2r + 0 = 116
   2r = 116
 1/2 * 2r = 1/2 * 116
          1r = 58
             r =58

Faster car average speed is 58mph
Slower car average speed is 51mph

Distance = rate * time
 d = 51 * 5 1/2
  d = 51 * 11/2
   d = 280.5


Slower car travel 280.5 miles in 5 1/2hrs
d = 58 * 5 1/2
d = 58 * 11/2
d = 319           or  599.5 - 280.5 = 319


Faster car travel 319 miles in 5 1/2 hrs




The slower car travels 280.5 miles in 5 1/2hrs. remainder of trips for both vehicle is 38.5 miles, because, 319 - 280.5 = 38.5


distance = rate * time
38.5 = 51 (t)
1/51(38.5) = 1/51(51)(t)
38.5/51 = 1t
77/102 = t, This time is in hours. To convert to minutes, multiply by 60 minutes per hour
77/102 * 60 = 77/51 * 30 = 2310/51 = 45, The slower car will arrive at about 45 minutes after the first car.



Hope this helps!!!


Next time please ask for help in chunks!! Try for yourself.. this is a lot...








5 0
2 years ago
Math<br><br><br><br> pls help!!<br><br><br><br><br><br> answers?
statuscvo [17]

Answer: Choice B) Infinitely many solutions

  • one solution: x = 8, y = -7/2, z = 0
  • another solution: x = -12, y = 13/2, z = 10

=======================================================

Explanation:

Here's the starting original augmented matrix.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\-4 & 0 & -8 & -32\\\end{array}\right]

We'll multiply everything in row 3 (abbreviated R3) by the value -1/4 or -0.25, which will make that -4 in the first column turn into a 1.

We use this notation to indicate what's going on: (-1/4)*R3 \to R3

That notation says "multiply everything in R3 by -1/4, then replace the old R3 with the new corresponding values".

So we have this next step:

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\1 & 0 & 2 & 8\\\end{array}\right]\begin{array}{l}  \ \\\ \\(-1/4)*R3 \to R3\\\end{array}

Notice that the new R3 is perfectly identical to R1.

So we can subtract rows R1 and R3, and replace R3 with the result of nothing but 0's

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\5 & 1 & 9 & 73/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\\ \\R3-R1 \to R3\\\end{array}

Whenever you get an entire row of 0's, it <u>always</u> means there are infinitely many solutions.

-------------------

Now let's handle the second row. That 5 needs to turn into a 0. We can multiply R1 by 5, and subtract that from R2.

So we need to compute 5*R1-R2 and have that replace R2.

\left[\begin{array}{ccc|c}  1 & 0 & 2 & 8\\0 & 1 & -1 & -7/2\\0 & 0 & 0 & 0\\\end{array}\right]\begin{array}{l}  \ \\5*R1-R2 \to R2\ \\\ \\\end{array}

Notice that in the third column of R2, we have 9-5*2 = 9-10 = -1. So we have -1 replace the 9. In the fourth column of R2, we have 73/2 - 5*8 = -7/2. So the -7/2 replaces the 73/2.

--------------------

At this point, the augmented matrix is in RREF form. RREF stands for Reduced Row Echelon Form. It seems a bit odd that the "F" of "RREF" stands for "form" even though we say "form" right after "RREF", but I digress.

Because the matrix is in RREF form, this means R1 and R2 lead to these equations:

R1 : 1x+0y+2z = 8\\ R2: 0z+1y-1z = -7/2

which simplify to

R1: x+2z = 8\\R2: y-z = -7/2

Let's get the z terms to each side like so:

x+2z = 8\\x = -2z+8\\\text{ and }\\y-z = -7/2\\y = z-7/2\\

Therefore, all of the solutions are of the form (x,y,z) = (-2z+8, z-7/2, z) where z is any real number.

If z is allowed to be any real number, then we can simply pick any number we want to replace it. We consider z to be the "free variable", in that it's free to be whatever it wants. The values of x and y will depend on what we pick for z.

So the concept of "infinitely many solutions" doesn't exactly mean we can pick just <em>any</em> triple for x,y,z (admittedly it would be nice to randomly pick any 3 numbers off the top of my head and be done right away). Instead, we can pick anything we want for z, and whatever we picked, will directly determine x and y. The x and y are locked into place so to speak.

Let's say we picked z = 0.

That would lead to...

x = -2z+8\\x = -2(0)+8\\x = 8\\\text{ and }\\y = z-7/2\\y = 0-7/2\\y = -7/2\\

So z = 0 would lead to x = 8 and y = -7/2

Rearranging the items in alphabetical order gets us:

x = 8, y = -7/2, z = 0

We have one solution of (x,y,z) = (8, -7/2, 0)

Now let's say we picked z = 10

x = -2z+8\\x = -2(10)+8\\x = -12\\\text{ and }\\y = z-7/2\\y = 10-7/2\\y = 13/2\\

So we have x = -12, y = -13/2, z = 10

Another solution is (x,y,z) = (-12, 13/2, 10)

There's nothing special about z = 0 or z = 10. You can pick any two real numbers you want for z. Just be sure to recalculate the x and y values of course.

To verify each solution, you'll need to plug them back into the original equations formed by the original augmented matrix. After simplifying, you should get the same thing on both sides.

8 0
2 years ago
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