Oh, dude, that's like a lot of x's and t's and some random letters thrown in there. So, technically, if you add all the x's, you get 7x, add all the t's, you get 3t, and the random letters are just there for decoration, I guess. So, the sum of x plus x plus x plus x plus x plus x plus x plus t plus t plus t plus jjakw is 7x + 3t + jjakw.
(t+h)(x+2)
(x + 2)(h + t)
45 + x + 54 + x 2x + 99
X 3 plus x-142 plus x is an equation equal to x3+2x-142.
What does the question ask? Does it ask you to simplify the above, or is it part of a larger question where values for t, x and h are given? Is it definitely Tx rather than tx? If asked to simplify it, there are numerous ways to do so (but none really make it 'simpler' as such, so it is a bit of a strange one). For example, by factorising: tx + 2t + hx + 2h = t(x + 2) + h(x+2) or we could write (again by factorising) tx + 2t + hx + 2h = x(t + h) + 2(t + h) Which is useful depends on what the question asks, really.
2 x t + 35 x t 2t + 35t 37t
2d+2h+2t+g+f+2r+w+x
3tx + 6t + 3hx +6h = 3(tx + 2t +hx +2h) = 3[t(x + 2) + h(x + 2)] = 3(x + 2)(t + h)
void swap(int &x, int &y) { x ^= y ^= x; } - or - void swap(int &x, int &y) { int t = x; x = y; y = t; }
(x + 2)(h + t)
(t+h)(x+2)
This equation has no meaning. 47 cubed is 103,823, but the rest of the equation is worthless without a value for x and t. At present, X and T do not have values, so you can not evaluate this expression.
x2 is the same as x times x. In this case x = t+2 so we can say (t+2)2 is (t+2)(t+2) or t2+4t+4
Use the following template function: template<class T> T& max(T& x, T& y){return(y<x?x:y;}
#include<iostream> template<typename T> void exch(T& a, T& b) { T temp = a; a = b; b = temp; } int main() { int x=4, y=2; exch(x,y); }
4x-4y-ty+xt = 4x+tx-4y-ty = x(4+t) - y(4+t) = (x-y)*(4+t)
T = 4x + 8 Subtract 8 from both sides: T - 8 = 4x Divide both sides by 4: T/4 - 2 = x or x = T/4 - 2