The given equation
x/2 = y/3 = z/4
can be broken into three separate equations which I'll call equations (A), (B) and (C)
- x/2 = y/3 ..... equation (A)
- y/3 = z/4 .... equation (B)
- x/2 = z/4 .... equation (C)
We'll start off solving for z in equation (C)
x/2 = z/4
4x = 2z ... cross multiply
2z = 4x
z = 4x/2 ... divide both sides by 2
z = 2x
Now let's solve for y in equation (A)
x/2 = y/3
3x = 2y
2y = 3x
y = 3x/2
y = (3/2)x
y = 1.5x
The results of z = 2x and y = 1.5x both have the right hand sides in terms of x. This will allow us to replace the variables y and z with something in terms of x, which means we'll have some overall expression with x only. The idea is that expression should simplify to 3 if we played our cards right.
We won't be using equation (B) at all.
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The key takeaway from the last section is that
Let's plug those items into the expression (2x-y+5z)/(3y-x) to get the following:
(2x-y+5z)/(3y-x)
(2x-y+5(2x))/(3y-x) ..... plug in z = 2x
(2x-y+10x)/(3y-x)
(12x-y)/(3y-x)
(12x-1.5x)/(3(1.5x)-x) .... plug in y = 1.5x
(12x-1.5x)/(4.5x-x)
(10.5x)/(3.5x)
(10.5)/(3.5)
3
We've shown that plugging z = 2x and y = 1.5x into the expression above simplifies to 3. Therefore, the equation (2x-y+5z)/(3y-x) = 3 is true when x/2 = y/3 = z/4. This concludes the proof.
malai kai tha xaina3521457269026422
Answer: 15cm^2
Step-by-step explanation:
0.5*10*3=15
Short answer: you don't.
The linear term in the numerator of the integral means the form shown is not applicable. Rather, you perform the integration using partial fraction expansion.
The integral is ...
... (1/35)ln|5x-1| +(6/35)ln|5x+13| +C
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If the numerator of your integral were a constant, then the fractions multiplying the separate partial fraction integrals would have the same magnitude and opposite signs. You would end with the difference of logarithms, which could be expressed as the log of a ratio as shown in your problem statement.
