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Even if we added the 1 and 2, first, we still have

6/2(3)  ...but the order of operations dictates that we perform multiplications and divisions at the same level of precedence as we encounter them from left to right....

And  since we encounter the division first, we have 6/2 = 3

And then

3(3) = 9

My advice??? ...Don't rely too much on those "intelligent friends and associates" for a lot of heavy-duty math guidance .... ( LOL..!!!)

This isn't possible.......both numbers have to be positive......and the largest possible result of multiplying two positive numbers that sum to 14 is 49.......that is, when x and y both = 7.

Exactly what I said but some very intelligent friends and associates feel as if it is worked out as 6/2(1+2)... 1+2=3 making it 6/2(3)....... 2(3)=6 making it 6/6...... which equals 1..... makes sense in certain situations but I don't believe this is the one lol

We can first count sets and then arrangements of those sets

We want to choose 1 of the 47 hip-hop MP3s and any 5 of the 14 Celtic Folk MP3s

The total number of sets of songs is given by C(47,1)* C(14,5) = 94094

But......each set of songs can be arranged in 6! = 720 ways

So   94094 x 720 = 67,747,680 different programs

A + 1/A  = 2    multiply everything by A

A^2 + 1 = 2A   subtract 2A from both sides

A^2 - 2a  + 1  = 0  factor

(A - 1) (A -1)  = 0     setting either one of the factors to 0, and we get that A =1

(5 -3) and (-9 -6)

1st...subtract the y's

-6 - (-3)  = -3

Then, subtract the x's in the same order

-9 - 5  = -14

Put the first result over the second

-3/-14   =  3/14     and that's the slope...!!!!

Let's break it down into prime factors

1282 = 2 x 641

So....it's just divisible by those two  things

"Of" means "multiply"     ..... so we have

8/9  x  774

Multiply 774 x 8  and divide this result by 9

774 x 8 = 6192

6192 / 9  =

688

And that's it....

2^-3 = 1/2³ = 1/8

{We are using the rule that  a^-n  = 1/aⁿ }

x² +x= 62     subtract 62 from each side

x² + x - 62 = 0   this doesn't factor......using the onsite solver, we have

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{62}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{249}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{249}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{8.389\: \!866\: \!919\: \!029\: \!75}}\\
{\mathtt{x}} = {\mathtt{7.389\: \!866\: \!919\: \!029\: \!75}}\\
\end{array} \right\}$$

Two real solutions......

Mmmmmmm....this doesn't seem to be too controversial, at all........

6/2(1+2)

3(1+2)  =

3(3)  =

9

We have that

54 = [(b1 + b2)/ 2] *h

54 = [(b1 + b2)/2] * 6

9 = (b1 + b2) / 2    ......and (b1 + b2) / 2 is the median......so, it equals 9

As an addendum to my above answer.... let's suppose that the 2 and 1 have to be together..........if we allowed for leading zeroes in both of the first two positions....we would have  300 possibilites.

If we didn't allow for a leading zero in the first position, we would have 100 + 90 + 90 = 280 possibilities.

E=IR

E=voltage in volts, I=current in amps, R=resistance in ohms

This isn't the fastest way to do it....but let's see if we can discover what the answer might be.....let's assume a number can't start with zero....I'm also assuming that the 2 and the 1 can be separated......this may or may not be the correct assumption....

Let's suppose that the 2 and 1 came first.....then we have 10 ways to fill the third posiition and 10 ways to fill the last position  = 100

And if the 1 is shifted to the right one position we again have 100 possibilities - 10 ways to fill the second position and 10 ways to fill the 4th position -  and if it's the last number that's 100 more.

So, when the 2 occupies the first position, we have 300 possibilities

Now, assuming that a number can't start with zero, let the 2 and 1 occupy the second and third positions. Then we have 9 x 10 = 90 ways to fill the other two positions - 9 ways for the first position times 10 ways for the last position. And this is true again if the 2 occupies the second position and the 1 occupies the last position.

And if the 2 and 1 occupy the last two positions, we again have 9 x 10 = 90 more possibilities.

So the total number would be 3x100 + 3x90 = 570

A number that multipied by its self to equal the said square root.

8C2 * 3!

$${\left({\frac{{\mathtt{8}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)}{\mathtt{\,\times\,}}{\mathtt{3}}{!} = {\mathtt{168}}$$

This allows a leading zero.

the 21 must stay as 21, the digits must be adjacent and the order CANNOT be reversed.

Ok....we know an opposite side - 10 -  and an adjacent side -14. We can find the angle of elevation  -  Θ - with the tangent inverse....so we have

tan^-1(10/14) = Θ = about 35.54° = 36°

2$${\mathtt{2}} = {\mathtt{2\times 10^{0}}}$$

a²⁺ b²⁼ c²

p=vi power equals voltage times current i is current p is watts and v is voltage