can hotelling's T square value be negative?
829736
what is "hotelling"? and is "T" a constant?\n\ni only know that any squared value is ALWAYS positive...
941569
yeah, i think so. but then put it into your caluclator and get the answer. (The neg value you ignore because obviously you cant sell them for -money)
973337
The minimum value is when x and y both equal zero.\n\nAny other value would be greater.  This is because any number, even if it is negative, squared gives a positive number.  Two positive numbers produce a positive sum.  Which means that the smallest number possible is 0, when both x and y are zero.\n\n\nHope this helps.
406488
If the number in front of x^2 is positive it opens up.\nIf it is negative it opens down.
960662
ry +s = tx - m\nYou must leave y alone for indicate its value in terms of other variables.\n\nFirst of all you move the s to the other side or the = as shown: ry =tx- m - s (you change the sign of s)\n\nThen you move the r as follows: y=(tx-m-s)/r\n\nAnd thats it \n\nYou have a restriction because the r has to be different than 0 to avoid a division by zero.  So restriction is r different from zero.
698011
The error is +/- 0.01.\n\nThe value then is: 6.02 +/- 0.01
461076
the value of the expression increases.\n\nUmm don't think so cuz negative numbers are real numbers and a negative number wouldnt make sense.
642997
The lim as x approached 7 is negative 2.\n\nThis would take my forever to type, I'll just explain. Plug in 7 and you get zero over zero. 0/0. Indeterminate form. \n\nYou can then apply L'hospital once, plug in 7 again, and you get -2.
733865
its between negative one and 0
1070381
It will be a zero because of the x-x term
415757
I hate to tell you, but I don't think there is an algebraic method for arriving at a value for x. Sometimes, this can be the case, in which case all you can do is find a numeric approximation, here around x=1.0587...\n\nI'll see if another user can prove me wrong, but I'm going to go with there's no way to solve this problem using analytic methods.
307198
It is linear on intervals, on (- infinite, 0) and on (0, +infinite) but not on the whole R. \n\nTry plotting it, you get a reversed V, not a straight line.
464388
pablodiablo is right, you can't treat an inequality like an equal sign when dividing or multiplying both sides by a negative number...swap the sign when doing so.  if you look at the graph of a logarithmic function, if 0<n<1, then log(n) is a negative number, and since y>x, x/y is less than one.  also, x & y must both be the same sign (positive or negative), otherwise x/y will be negative, and log of a negative number is undefined.
212459
If you were to multiply it out, your x^2 term will be negative (you don't need to multiply it out, just look to see whether it is + or -), so it would have a maximum and face down.
968535
first-negative 28x\n\nsecond- negative 29.4x
242385
when you multiply two negatives you get a positive because the negative signs cancell each other out
468001
x+2 must not be 0\nso x must not be -2\ndomain is \nall real numbers - (-2)\nall real no except -2