Mister Exam
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How to use it?
- Equation:
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Equation 15-x=8
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Equation x^2+3*x+9=0
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Equation x^2-16*x+64=0
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Equation x^2-6x-7=0
- Express {x} in terms of y where:
- 9*x-8*y=-6
- -11*x+1*y=-14
- -13*x-14*y=0
- 7*x-17*y=2
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Identical expressions
- eighty-four = zero , seven * four thousand, two hundred * zero , three / twenty-one *log(two) sixty - twenty /x- twenty
- 84 equally 0,7 multiply by 4200 multiply by 0,3 divide by 21 multiply by logarithm of (2)60 minus 20 divide by x minus 20
- eighty minus four equally zero , seven multiply by four thousand, two hundred multiply by zero , three divide by twenty minus one multiply by logarithm of (two) sixty minus twenty divide by x minus twenty
- 84=0,742000,3/21log(2)60-20/x-20
- 84=0,742000,3/21log260-20/x-20
- 84=O,7*4200*0,3/21*log(2)60-20/x-20
- 84=0,7*4200*0,3 divide by 21*log(2)60-20 divide by x-20
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Similar expressions
- 84=0,7*4200*0,3/21*log(2)60-20/x+20
- 8(4-3x)+7(x-3)+3(9+7x)=10
- 84=0,7*4200*0,3/21*log(2)60+20/x-20
84=0,7*4200*0,3/21*log(2)60-20/x-20 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
1 20
84 = 7/10*4200*3/10*--*log(2)*60 - -- - 20
21 x
$$84 = \left(-1\right) 20 + \frac{7}{10} \cdot 4200 \cdot \frac{3}{10} \cdot \frac{1}{21} \log{\left(2 \right)} 60 - \frac{20}{x}$$
Detail solution
Given the equation:
$$84 = \left(-1\right) 20 + \frac{7}{10} \cdot 4200 \cdot \frac{3}{10} \cdot \frac{1}{21} \log{\left(2 \right)} 60 - \frac{20}{x}$$
Use proportions rule:
From $\frac{a_1}{b1} = \frac{a_2}{b_2}$ should $a_1*b_2 = a_2*b_1$,
In this case
so we get the equation
$$1 \left(- \frac{x}{20}\right) = 1 \cdot \frac{1}{- 2520 \log{\left(2 \right)} + 104}$$
$$- \frac{x}{20} = \frac{1}{- 2520 \log{\left(2 \right)} + 104}$$
Expand brackets in the right part
Divide both parts of the equation by -1/20
We get the answer: x = 5/(2*(-13 + 315*log(2)))
$$84 = \left(-1\right) 20 + \frac{7}{10} \cdot 4200 \cdot \frac{3}{10} \cdot \frac{1}{21} \log{\left(2 \right)} 60 - \frac{20}{x}$$
Use proportions rule:
From $\frac{a_1}{b1} = \frac{a_2}{b_2}$ should $a_1*b_2 = a_2*b_1$,
In this case
a1 = 1
b1 = 1/(104 - 2520*log(2))
a2 = 1
b2 = -x/20
so we get the equation
$$1 \left(- \frac{x}{20}\right) = 1 \cdot \frac{1}{- 2520 \log{\left(2 \right)} + 104}$$
$$- \frac{x}{20} = \frac{1}{- 2520 \log{\left(2 \right)} + 104}$$
Expand brackets in the right part
-x/20 = 1/104+1/2520*log+1/2)
Divide both parts of the equation by -1/20
x = 1/(104 - 2520*log(2)) / (-1/20)
We get the answer: x = 5/(2*(-13 + 315*log(2)))
Rapid solution
[src]
5
x_1 = --------------------
2*(-13 + 315*log(2))
$$x_{1} = \frac{5}{2 \left(-13 + 315 \log{\left(2 \right)}\right)}$$
Sum and product of roots
[src]
sum
5 -------------------- 2*(-13 + 315*log(2))
$$\left(\frac{5}{2 \left(-13 + 315 \log{\left(2 \right)}\right)}\right)$$
=
5 -------------------- 2*(-13 + 315*log(2))
$$\frac{5}{2 \left(-13 + 315 \log{\left(2 \right)}\right)}$$
product
5 -------------------- 2*(-13 + 315*log(2))
$$\left(\frac{5}{2 \left(-13 + 315 \log{\left(2 \right)}\right)}\right)$$
=
5 -------------------- 2*(-13 + 315*log(2))
$$\frac{5}{2 \left(-13 + 315 \log{\left(2 \right)}\right)}$$
The graph