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the third one was supposed to be f times g not minus.

thanks for your help though

Yes.

Seven would not be possible when the final amount is only THREE gallons.

You WEIGH less on the moon.....approx 1/6 of your Earth weight....HOWEVER : YOUR MASS remains the same!

\(\bar{\Delta}_{[7,9]} = \dfrac{3878-1852}{9-7} = 1013\\ \bar{\Delta}_{[4,6]} = \dfrac{1178-358}{6-4}=410\\ \bar{\Delta}_{[7,9]}-\bar{\Delta}_{[4,6]} = 1013-410 = 603\)

\(f(x) = x^3-x\\ g(x) = x^3+2x^2-10\\ (f+g)(x) = f(x) + g(x) = 2x^3 +2x^2 - x - 10\)

\(f(x) = x^2+2x-6\\ g(x) = x^4+3\\ (f-g)(x) = f(x)-g(x) = -x^4+x^2+2x-9\)

If        \(\displaystyle \frac{2}{a}-\frac{3}{b}=\frac{1}{5},\)

then

\(\displaystyle 10b-15a=ab,\)

so

\(\displaystyle b =\frac{15a}{10-a}.\)

\(\displaystyle \text{Both }a\text{ and } b\text{ are to be positive integers,}\)

so there are just nine possible values for \(\displaystyle a ,\;\;\; 1,2,\dots,9.\)

Just check each one in turn.

You have to take the LEAST costly item, which is the "spin top" and sell it as follows:

x% * $1.50 =$2.50  divide both sides by 1.50

x% =2.50 / 1.5 =1.6667 - 1 x 100 =66.67% - is the minimum percetage markup in order to earn at leas $1 in profit on every item in your list.

Do the same as your other question:

$12 x 1.20 =$14.40 - sale price after 20% markup.

$14.40 - $12.00 =$2.40 - profit on each wooden boat.

$15 / 2 = $7.50 - the cost of  one necklace.

$7.50 x 1.15 =$8.63 - sale price of each necklace after 15% markup.

$8.63 - $7.50 =$1.13 - profit on each necklace.

Look here under Brent Follin and see the "Equation of Life, the Universe and everything in it"!!!!

https://www.quora.com/What-is-the-best-all-encompassing-equation-for-physics-that-we-have-now

Well Rom, I agree with you and Melody, but if W/A and Mathematica 11 Home edition for that matter, don't give ANY explanations as to how they derive these numbers, then all I can think of is that they must use iteration, or trial and error essentially, to get these numbers!!. Since I have Newton-Raphson iteration and interpolation method programmed into my computer, I can essentially do the same thing they do.

Sorry, that is the only explanation I have.

\(happiness \sim |tacos|\)

flip a and b and they all work

but I agree a proper answer would have at least tried to show how those pairs were derived.

135,67

Yes but the asker wants YOU to explain how to find them.

I mean WITHOUT simply referring to Wolfram Alpha.

There are 5 pairs of POSITIVE integers that satisfy the equation. See here:

https://www.wolframalpha.com/input/?i=2%2Fa+-+3%2Fb+%3D1%2F5,+iterate+for+a,+b

See the answer here:  https://web2.0calc.com/questions/math-postage-stamp-problem

Azsan,    5.5 means 5*5      

I can understand your confusion but the 2 notations are interchangable. 

Guest's answer is excellent.

\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x = -1+2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x +1= 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x +1= 2 + \cfrac{1}{x+1}\\ \)

\(x +1= 2 + \cfrac{1}{x+1}\\ x-1= \cfrac{1}{x+1}\\ x^2-1=1\\ x^2=2\\ x=\pm\sqrt2 \)

I do not know why the answer cannot be -sqrt2  but if memory serves me correctly the only correct answer is +sqrt2

I'll continure thinking about that one.

\(\frac{2}{a} - \frac{3}{b} = \frac{1}{5}\\\)

Lets look at Partial Mathemticians answers:  

They are (10, 4), (15, 5), (35, 7), (60, 8), and (135, 9).

(10, 4)    2/10-3/4  not equal      1/5    doesn't work,

(15, 5),      2/15 -3/5 =-7/15                 doesn't work

(35, 7),       2/35-3/7 = -13/35              doesn't work

(60, 8),        2/60 - 3/8 = -41/120          doesn't work

and (135, 9).     2/135 -3/9 =  -43/135   doesn't work

None of his answers work, maybe the a and b are the wrong way around. I have not looked at that.

No one should asssume that any answer is correct.   It is up to the question asker to check.

\(\frac{2}{a} - \frac{3}{b} = \frac{1}{5}\\ 5ab(\frac{2}{a} - \frac{3}{b})= 5ab(\frac{1}{5})\\ 10b-15a=ab\\ \)

Here is another modular equation. What is the remainder of: 17^1507 mod 102? Will appreciate any help. Thank you.

102=6*17

17^1507 = 0 (mod17)

so  17^1507 (mod 102)  will be   on of these.    0,17,34,51,68 or 85 

17^1507 (mod 6) = (-1)^1507 (mod 6) = -1

So which of the above possibilities will equal -1 (or 5 which is the same) in Mod 6

The only answer that works is 17

Can you walk me through the steps? I only came up with two ordered pairs (4,10) and (5,15)

They are (10, 4), (15, 5), (35, 7), (60, 8), and (135, 9).

5 pairs.

 - PM