If P1 200 times 1065 solve these three simultaneous equatio

If P_1 = 2.00 times 1065, solve these three simultaneous equations for P-2: T_2 = 3T_1 P_1 V_1^5/3 = P_2 V_2^5/3 P_1 V_1/T_1 = P_2 V_2/T_2 Recall that taking the nth root of a number is the same as raising that number to the l/n power. [And note that AB^C = A(B^C), not (AB)^C] (a) Choose any two different non-zero numbers. Show that the mean (average) of their squares is not always equal to the square of their mean. (b) Using just two variable names (such as n and b). Prove that the above fact is true in general - that the square of the mean is not always equal to the mean of the square. The length of a rectangular sheet of metal decreases by 34.5 cm. Its width decreases proportionally -that is, by the same percentage. If the sheet\'s original width was half the original length, and that new (smaller) are of the sheet is 1.20 m^., (a) what was the sheet\'s original width? The length of a rectangular sheet of metal decreases by 34.5 cm. Its width decreases proportionally -that is, by the same percentage. If the sheet\'s original width was half the original length, and that new (smaller) are of the sheet is 1.20 m^., (a) what was the sheet\'s original width?

Solution

Solution : ( 5 )

First rearrange the second equation :

P₁*(V₁^(5/3)) = P₂*(V₂^(5/3))
[P₁*(V₁^(5/3))]/P₂ = V₂^(5/3)
V₂ = [(P₁^(3/5))*V₁]/[P₂^(3/5)]

Replace T₂ by 3T₁ in the third equation :
(P₁*V₁)/T₁ = (P₂*V₂)/T₂
(P₁*V₁)/T₁ = (P₂*V₂)/(3T₁)

And rearrange :
V₂ = (3T₁*P₁*V₁)/(T₁*P₂)
the T1\'s cancel :
V₂ = (3*P₁*V₁)/P₂

Since both the second and third equations equal V₂, they equal each other:
[(P₁^(3/5))*V₁]/[P₂^(3/5)] = (3P₁*V₁)/P₂

Rearrange and solve for P₂:
P₂/(P₂^(3/5)) = (3P₁*V₁)/[(P₁^(3/5))*V₁]
...the P₂ and P₁ can be simplified and the V₁ cancel (or if you prefer you can plug in the value for P₁ at this point)
P₂^(2/5) = (6*10⁵)/[(2*10⁵)^(3/5)]
P₂ = 3.118*10⁶

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Solution:

(a)

Let original width = x

Original length = 2x

Original area = 2x²

New length = 2x - 0.345

% decrease in length = 0.345 * 100/2x = 34.5/2x

% change in width = (34.5/2x)(x) = 17.25

New width = x – 0.1725

New area = (2x - 0.345)(x – 0.1725) = 2x² – 0.69x + 0.06

By data, 2x² – 0.69x + 0.06 = 1.2

On solving, we get x = 0.947 m as the original width.