The first step was the combine the like terms; in this example he is subtracting the 3x on both sides of the =.
If you were to solve the equation it would look like this:
2x+6=3x-8
-3x -3x
-x+6=-8
-6 -6
-x=-14
/-1 /-1
x=14
Answer:
The quadratic mean of 2 real positive numbers is greater than or equal to the arithmetic mean.
Step-by-step explanation:
x and y Quadratic Mean Arithmetic mean
3 and 3 3 3
2 and 3 2.55 2.5
3 and 6 4.74 4.5
2 and 5 3.8 3.5
2 and 17 12.1 9.5
18 and 28 23.5 23
10 and 48 34.7 29
The quadratic mean is always greater than the arithmetic mean except when x and y are the same.
When the difference between the pairs is small the difference in the means is also small. As that difference increases the difference in the means also increases.
So we conjecture that the quadratic mean is always greater than or equal to the arithmetic mean.
Proof.
Suppose it is true then:
√(x^2 + y^2) / 2) ≥ (x + y)/2 Squaring both sides:
(x ^2 + y^2) / 2 ≥ (x + y)^2 / 4 Multiply through by 4:
2x^2 +2y^2 ≥ (x + y)^2
2x^2 +2y^2 >= x^2 + 2xy + y^2
x^2 + y^2 >= 2xy.
x^2 - 2xy + y^2 ≥ 0
(x - y)^2 ≥ 0
This is true because the square of any real number is positive so the original inequality must also be true.
Answer:
just take the 96 and 66 and middle it then take y multiply that=answer
Step-by-step explanation:
the idea behind the recurring decimal as a fraction, is to first off, multiply or divide by some power of 10, in order that we leave the recurring decimal to the right of the decimal point.
then we multiply by a power of 10, in order to move the repeating digits to the left of the decimal point, anyhow, let's proceed.
![\bf 0.6\overline{1212}\implies \cfrac{06.\overline{1212}}{10}\implies \cfrac{6+0.\overline{1212}}{10}\qquad \qquad \stackrel{\textit{now let's make}}{x=0.\overline{1212}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{llll} 100\cdot x &=& 12.\overline{1212}\\\\ &&12+0.\overline{1212}\\\\ &&12+x\\\\ 100x&=&12+x\\\\ 99x&=&12\\\\ x&=&\cfrac{12}{99}\implies x = \cfrac{4}{33} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%200.6%5Coverline%7B1212%7D%5Cimplies%20%5Ccfrac%7B06.%5Coverline%7B1212%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2B0.%5Coverline%7B1212%7D%7D%7B10%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bnow%20let%27s%20make%7D%7D%7Bx%3D0.%5Coverline%7B1212%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20100%5Ccdot%20x%20%26%3D%26%2012.%5Coverline%7B1212%7D%5C%5C%5C%5C%20%26%2612%2B0.%5Coverline%7B1212%7D%5C%5C%5C%5C%20%26%2612%2Bx%5C%5C%5C%5C%20100x%26%3D%2612%2Bx%5C%5C%5C%5C%2099x%26%3D%2612%5C%5C%5C%5C%20x%26%3D%26%5Ccfrac%7B12%7D%7B99%7D%5Cimplies%20x%20%3D%20%5Ccfrac%7B4%7D%7B33%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \cfrac{06.\overline{1212}}{10}\implies \cfrac{6+x}{10}\implies \cfrac{6+\frac{4}{33}}{10}\implies \cfrac{~~\frac{202}{33}~~}{10}\implies \cfrac{~~\frac{202}{33}~~}{\frac{10}{1}}\implies \cfrac{202}{330} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{101}{165}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B06.%5Coverline%7B1212%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2Bx%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2B%5Cfrac%7B4%7D%7B33%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B202%7D%7B33%7D~~%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B202%7D%7B33%7D~~%7D%7B%5Cfrac%7B10%7D%7B1%7D%7D%5Cimplies%20%5Ccfrac%7B202%7D%7B330%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Ccfrac%7B101%7D%7B165%7D~%5Chfill)
notice, we first divided by 10, to move the decimal point over to the right by 1 slot, then we multiplied by 100, to move it two digits over the decimal point, namely the repeating "12", thus we use 100.
By using the definition of inverse functions, we will see that:
g(g(f(f(f(36))))) = f(36) = 25.
<h3>
What are inverse functions?</h3>
Two functions f(x) and g(x) are inverses if:
f( g(x) ) = x
g( f(x) ) = x
Then we can rewrite:
g(g(f(f(f(36)))))
First, we can see that:
g(f(f(f(36)))) = f(f(36))
Replacing that in our expression, we get:
g(g(f(f(f36))))) = g(f(f(36)))
And the above expression is equal to f(36), to be sure of that, let's replace:
u = f(36)
Then we can rewrite:
g(f(f(36))) = g(f(u))
And by definition, the above thing is equal to u:
g(f(u)) = u = f(36).
Finally, we conclude that:
g(g(f(f(f(36))))) = f(36) = 3*√36 + 7 = 3*6 + 7 = 25
If you want to learn more about inverse functions, you can read:
brainly.com/question/12220962
