i solve this when i was in grade 11.?
405393
3, 3, 8.  That's the only combination that has another combination that adds to the same value (2,6,6) so that the guest couldn't know the ages just by looking at the house number (14).  And of the two, this one has an oldest.
658148
given 4a^2-12ab+9b^2\n    =(2a)^2-2(2a)(3b)+(3a)^2\n    =(2a-3b)^2
1111502
okay here is what I think....still a little unclear\nx=9th grade\nx+8=8th grade\n3/4(x+8)=11th grade\n\nso...\nx+ x+8+ 3/4(x+8)=96\n\nx+x+8+3/4x+6=96\ncombine terms\n2 3/4x + 14=96\nsubtract 14 from both sides\n2 3/4x=82\ndivide each side by 2 3/4\nx=29 9/11\nthen round up\n9th grade =30\n8th grade =38\n11th grade=28 (actually it is28 1/2 but bc you can't have 1/2 of a person it is 28)
810076
11^(0.35+7) = 3 this is a contradiction\n\nx=3 is a solution but the rest is just a mess.\n\nDo you have any idea of what you are doing?
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
1089449
It's as simple as plugging in numbers.  The number 15 becomes the letter "a", the number 7 is "b", and the number 55 is "c".  Just substitute the number for the letter in the formula and solve.
636051
x=4\n\nplease solve...\n\n8-(1) :-) ;-(
920569
It looks like you are trying to solve a literal equation, which is an equation with more than one variable, and you have to solve for a specified variable.  You can use the SAME rules to solve this kind of equation as you do with regular equations (those with numbers).  For example, if you had 8=2(b)(9), you would divide by 2 and then again by 9 (which is like dividing by 18).  \n\nIf your problem is a=(1/2)bc^2, then you would divide by 1/2 and then again by c^2.  Remember that to divide by a fraction, you multiply by the reciprocal.
857682
The midpoint of A and B is (15+7)/2 = 11.\nIf C = 11 then AC =BC.\nTherefore AC < BC  for 11<C<15
137681
Ok...\nI see what you are doing...\nYou merely substitute the number in for x...\nso the problem becomes 6-2*(-1)\n6+2\n8
966925
There are 3 solutions for ABCDEFGHJ -\n\n2 6 7 8 3 5 1 4 9\n2 6 7 9 5 3 1 4 8\n6 7 8 4 1 3 2 5 9
817481
We can use faux-algebra to solve these. You probably haven't taken algebra yet. With algebra these would be extremely easy to solve. It's why algebra was invented.\n\na. 16 x ___ has to be -42, because if we add 51 to it we get 9. Thus, ____ has to be -42/16, or -21/8.\n\nb. 2 x ____ has to be 22 - 9 x _____, because if we subtract 8 from it we get 14 - 9 x ____. And 22 has to be 11 x ____, because if we subtract 9 x ____ from it we get 2 x ____. Thus, ____ has to be 22/11, or 2.\n\nc. 9 x ___ has to be 3 x ____ + 3, because if we add 2 to it we get 3 x ___ + 5.  And 3 has to be 6 x ____, becuase if we add 3 x ____ to it we get 9 x ____. Thus, ____ has to be 3/6, or 1/2.\n\nWith algebra, no such thinking would be required. You just plug in the formulas and grind away. You'll see when you study algebra.
959197
I have factored it please refer to.\n\nFrom this solution can be found
799993
Use the following to help solve this problem:\n\ni ^1 = i\ni ^2 = -1\ni ^3 = -i\ni ^4 = 1\n\nThis pattern then repeats:\n\ni ^5 = i^4 * i = 1 * i = i\ni ^6 = i^4 * i^2 = 1 * -1 = -1\n\netc.\n\nSo to solve any i ^n, just look at the remainder of n/4:\n\nremainder \nof n/4 ....... value of i ^n\n- - - -  - - - - - - - - - - - - - - -\n0 ....................... 1\n1 ........................ i\n2 ....................... -1\n3 ........................ -i\n\nSo now let's tackle your specific problem:\n\ni ^16 = 1 (since n=16, n/4 = 4 with remainder of 0)\ni ^6 = -1 (since n=6, n/4 = 1 with remainder of 2)\ni ^5 = i (since n=5, n/5 = 1 with remainder of 1)\ni ^13 = i (since n=13, n/4 = 3 with remainder of 1)\n\nSo, i^16 + i^6 - 2i^5 + i^13 = \n1 + (-1) - 2(i) + i =\n-i\n\nHope that helped!\n\n~ ♥ ~\n\nPS -- The "2-i" answers are incorrect.  Those answerers just made a tiny typo and took 1 - (-1) as the first two terms, rather than 1 + (-1).  Just wanted to let you know, so that you don't put the wrong answer down on your homework.
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
781088
Plug in 2 for x to determine this:\n\n3(2) + 5 = 6+5 = 11 ≠ 17\n\nTherefore 2 is not a solution.\n\n~ ♥ ~
571233
Curious Georg. gave it to you perfectly. If you don't know how to do this very simple calculation, you should talk to your teacher or you parents, because you'll fall way behind.