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Vlad [161]
3 years ago
9

Volume equals length times width times height v=lwh solve the equation for l

Mathematics
1 answer:
marshall27 [118]3 years ago
3 0

Answer:

\frac{V}{wh}=l

Step-by-step explanation:

Equation: V=lwh

1). \frac{V}{wh}=\frac{lwh}{wh}

2). \frac{V}{wh}=l

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Do oz so oz.  Top oz so off.

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What is the value of x in the equation X=-45?<br> O-75<br> O -27<br> O 27<br> O 75
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Step-by-step explanation:

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Three dice are rolled. Let the random variable x represent the sum of the 3 dice. By assuming that each of the 63 possible outco
tamaranim1 [39]
<h3>Answer:  1/8</h3>

In decimal form, 1/8 = 0.125 which converts to 12.5%

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

Work Shown:

The 63 should be 6^3. There are 6 choices per slot, and 3 slots, so 6^3 = 216 different outcomes.

Here are all of the ways to add to 11 if we had 3 dice

  1. sum = 1+4+6 = 11
  2. sum = 1+5+5 = 11
  3. sum = 1+6+4 = 11
  4. sum = 2+3+6 = 11
  5. sum = 2+4+5 = 11
  6. sum = 2+5+4 = 11
  7. sum = 2+6+3 = 11
  8. sum = 3+2+6 = 11
  9. sum = 3+3+5 = 11
  10. sum = 3+4+4 = 11
  11. sum = 3+5+3 = 11
  12. sum = 3+6+2 = 11
  13. sum = 4+1+6 = 11
  14. sum = 4+2+5 = 11
  15. sum = 4+3+4 = 11
  16. sum = 4+4+3 = 11
  17. sum = 4+5+2 = 11
  18. sum = 4+6+1 = 11
  19. sum = 5+1+5 = 11
  20. sum = 5+2+4 = 11
  21. sum = 5+3+3 = 11
  22. sum = 5+4+2 = 11
  23. sum = 5+5+1 = 11
  24. sum = 6+1+4 = 11
  25. sum = 6+2+3 = 11
  26. sum = 6+3+2 = 11
  27. sum = 6+4+1 = 11

There are 27 ways to add to 11  using 3 dice. This is out of 216 total outcomes of 3 dice being rolled.

So, 27/216 = (1*27)/(8*27) = 1/8 is the probability of getting 3 dice to add to 11.

7 0
3 years ago
(1.1/1.2: Interpolating polynomials) Say we want to find a polynomialf(x) ofdegree 3,f(x) =a0+a1x+a2x2+a3x3,satisfying some inte
Hunter-Best [27]

(a) If

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then from the given conditions we get the system of equations,

<em>f</em> (-1) = <em>a</em>₀ - <em>a</em>₁<em> </em>+<em> a</em>₂ - <em>a</em>₃ = -1

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+ 4<em>a</em>₂ + 8<em>a</em>₃ = 1

<em>f</em> (3) = <em>a</em>₀ + 3<em>a</em>₁<em> </em>+<em> </em>9<em>a</em>₂ + 27<em>x</em> ³ = 5

(b) Similarly, if

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then

<em>f'(x)</em> = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ <em>x</em> + 3<em>a</em>₃ <em>x</em> ²

so that the given conditions yield the system,

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 0

<em>f'</em> (1) = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ + 3<em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 27<em>a</em>₃ = 3

<em>f'</em> (2) = <em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 12<em>a</em>₃ = -1

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3 years ago
given the rule s=t+3 and the starting number 0, create a table to show the first 5 terms in the sequence graph the resulting ord
IRINA_888 [86]
0= 3
1= 4
2=5
3=6
4=7
5=8
6=9
7=10
8 0
3 years ago
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