It is: x(x+11) or x^2 +11x
It is an equation: 3x+11=32 and the value of x is 7
x times the sum of x and 9 x times (x times 9) x(9x) 9 times x[squared] 9x[squared]
First sum the terms with x: 9x + (-3)x = 6 x. Then sum the terms without x: -11 + (-13)x = -24. Then 6x - 24 =0 or x = 24/6 = 4
5(x^2 + y)
To find two times the sum of ( x^2 ) and ( y^2 ) increased by three times the sum of ( x^2 ) and ( y^2 ), we first express it mathematically. The sum of ( x^2 ) and ( y^2 ) is ( x^2 + y^2 ). Thus, two times this sum is ( 2(x^2 + y^2) ), and three times it is ( 3(x^2 + y^2) ). Adding these together gives ( 2(x^2 + y^2) + 3(x^2 + y^2) = 5(x^2 + y^2) ).
It is an equation: 3x+11=32 and the value of x is 7
The sum of a number and 11 can be represented as x + 11, where x is the unknown number. This expression denotes adding 11 to the unknown number. If you know the value of x, you can simply add 11 to find the sum. If x is unknown, the expression x + 11 remains as an algebraic representation of the sum.
3x+11 = 32 3x = 32-11 3x = 21 The solution is: x = 7
x times the sum of x and 9 x times (x times 9) x(9x) 9 times x[squared] 9x[squared]
(11 x 5) + (11 x 6) = 11 x 11 = 121
First sum the terms with x: 9x + (-3)x = 6 x. Then sum the terms without x: -11 + (-13)x = -24. Then 6x - 24 =0 or x = 24/6 = 4
5(x^2 + y)
11(5 + 6) = (11 x 5) + (11 x 6) = 11 x 11 = 121
x(9+x)=x^2 + 9x
11 x 11 x 11 = 1,331
2 x 3 x 7 x 11 = 6 x 77 = 462
=7(x+s)