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Alexxx [7]
2 years ago
15

3(x+3/4+5-x/4-x/6)=10​

Mathematics
1 answer:
viva [34]2 years ago
5 0

3(x+3-4+5-x-4-x-6)=10

-3x/-3= -28/3

x= -28/3

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16 cm - 2.5 cm = 13.5 cm, which is 1/4 of the diameter of the large gear. Take 13.5 cm * 4, and you have the diameter of the large gear, which is 54 cm.
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Solve for x <br>x/6=10<br>1) x=10<br>2) x=60<br>3) x=4<br>4) x=16​
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Q + 5/9= 1/6 from big ideas mathbook
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9We have such eqation
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2 years ago
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Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3
Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

\frac{dV}{dt} = -kA

Using Euler's method

\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i}    \\

Where initial droplet volume is:

V(0) = \frac{4pi}{3} * r(0)^3 =  \frac{4pi}{3} * 2.5^3 = 65.45 mm^3

Hence, the iterative solution will be as next:

  • i = 1, ti = 0, Vi = 65.45

V'_{i}  = -k *4pi*(\frac{3*65.45}{4pi})^(2/3)  = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88

  • i = 2, ti = 0.5, Vi = 63.88

V'_{i}  = -k *4pi*(\frac{3*63.88}{4pi})^(2/3)  = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33

  • i = 3, ti = 1, Vi = 62.33

V'_{i}  = -k *4pi*(\frac{3*62.33}{4pi})^(2/3)  = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\

The average change of droplet radius with time is:

Δr/Δt = \frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0
3 years ago
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