Interesting math puzzle.
I'm getting x^2 - 3x + 2.
Well, yes....there is only one. Which one? Substitute x = 2 and then x = 6 into the original equation to see which one works.
there is only one answer i think
Total candies = 16
First red 4/16
Now there are only 3 red out of 15
Second red = 3/15
Third red = 2/14
4/16 * 3/15 * 2/14 = ??
sqrt(2x-3) = x-3
2x-3 = (x-3)^2
2x-3 = x^2-6x+9
x^2-8x+12 = 0
(x-6)(x-2) = 0 so x = 2 or 6 Do both answers solve the equation? Or is one of them extraneous? What do YOU think?
No, you are not correct. The correct answer is x=6.
sum(1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , 28 , 31 , 34 , 37 , 40 , 43 , 46 , 49 , 52 , 55 , 58 , 61 , 64 , 67 , 70 , 73 , 76 , 79 , 82 , 85 , 88 , 91 , 94 , 97) =1,617
n = 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 , 25 , 28 , 31 , 34 , 37 , 40 , 43 , 46 , 49 , 52 , 55 , 58 , 61 , 64 , 67 , 70 , 73 , 76 , 79 , 82 , 85 , 88 , 91 , 94 , 97 , Total = 33 such numbers.
This is the answer assuming the first case I stated (2 different terms being multiplied)
\(\frac{\left(x^2+x-12\right)\left(3x^2+11x-4\right)}{\left(x-2\right)\left(x^2-4\right)}\)
Are you saying that (x ^ 2 + x - 12)/(x - 2) is getting multiplied by (3x ^ 2 + 11x - 4)/(x ^ 2 - 4) as 2 seperate terms, like (x ^ 2 + x - 12)/(x - 2) * (3x ^ 2 + 11x - 4)/(x ^ 2 - 4), or is (3x ^ 2 + 11x - 4)/(x ^ 2 - 4) in the denominator of (x ^ 2 + x - 12)/(x - 2)? Try using LaTeX to make the problem clearer.
Are these homework problems?
Using the log base change rule to change to log10 :
log x / log3 = (-2 + log100/log2)(log(sqrt2)/log3)
Now, using calculator you will find x = 5
Hmm. We seem to be disagreeing. I think you forgot that y = 2 + 5/6x is a different term than 4x-3y=3, lol. Guest probably should have put something to tell us that they are 2 different terms, unless I am the mistaken one. Oh I see you didn't see that it was a system of equations so you just solved for x!
As for the 5/6x confusion you are having, just put the 5/6x in the term you are substituting in for "y", like I did when I said \(\mathrm{Subsititute\:}y=2+\frac{5}{6}\cdot \:x\)
y = 2 + 20/6 x - 3y add 3 y to both sides of the equation
4y = 2 + (20/6) x your original Q says y = 3 substitute that in
12 = 2 + 20/6 x subtract 2 from both sides
10 = 20/6 x multiply both sides by 6/20
60/20 = x = 3
Now I see your question has TWO Equations....please put some spaces in there !
y = 2 + 5/6 * x
4x - 3y = 3 Is much more clear.....see whoisjoe answer below....
\(\mathrm{Subsititute\:}y=2+\frac{5}{6}\cdot \:x \)
\(\begin{bmatrix}4x-3\left(2+\frac{5}{6}x\right)=3\end{bmatrix}\)
\(\mathrm{Isolate}\:x\:\mathrm{for}\:4x-3\left(2+\frac{5}{6}x\right)=3:\quad x=6\)
\(\mathrm{For\:}y=2+\frac{5}{6}\cdot \:x\)
\(\mathrm{Subsititute\:}x=6\)
\(2+\frac{5}{6}\cdot \:6=7\)
\(y=7, \text{Solution is x=6, y=7}\)
\(r=\sqrt{\frac{f}{n+k-a}}, r=-\sqrt{\frac{f}{n+k-a}};\quad \:a\ne \:n+k\)
f/(r ^ 2) = n + k - a after subtracting 'a' from both sides of the equation
f/(n+k-a) = r^2 after re-arranging now take sqrt of both sides
r = +- sqrt ( f/(n+k-a) ) (and as WIJ pointed out n+k-a cannot equal 0)
... okay...
please try not to post topics like this, assuredbats. web2.0calc is a place where you ask any questions about topics, not make statements and post random things.
This is a spammer.
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thoughtco.com
If you can find very dark skies with no moon, you can see an ARM of the spiral Milky Way stretching from north to south across the sky....looks kinda like a thin high sirrus cloud....here is a couple of pics from last Summer: (one even has a faint meteor streaking across it)
You can solve for y in the first equation, to get y = (x^3 - 20x)/7.
You can then substitute into the second equations, which gives you
\(\left( \frac{x^3 - 20x}{7} \right)^3 = 7x + 20 \left( \frac{x^3 - 20x}{7} \right)\)
This factors as x(x - 5)(x + 5)(x^2 + 12)(x^4 - 20x^2 + 49) = 0. You can then work out the rest.
thanks!
