7x + 3 t +jjakw
(t+h)(x+2)
(x + 2)(h + t)
The answer is 5x+3y. (5 xs and 3 ys)
x+x+x= 3x
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
3tx + 6t + 3hx +6h = 3(tx + 2t +hx +2h) = 3[t(x + 2) + h(x + 2)] = 3(x + 2)(t + h)
2d+2h+2t+g+f+2r+w+x
void swap(int &x, int &y) { x ^= y ^= x; } - or - void swap(int &x, int &y) { int t = x; x = y; y = t; }
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.
(x + 2)(h + t)
(t+h)(x+2)
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