in that q should be >.
Explanation: intially replace q with <,then equation is,
3 < 11 + 8 < 99
3 < 19 < 99
Its Satisfying the condition so
as per ineauality it should be greater that (>).
so answer is replace greater than with Q
* * * * *
I think the above answer misses the point entirely. The question is intended to be an equation (or inequality) in q to be solved. However, thanks to limitations of the browser, symbols are not visible so the equation appears to make no sense.
3q+11+8q>99
To solve the equation ( 11 + 8q = 3q - 19 ), first, isolate the variable ( q ). Subtract ( 3q ) from both sides to get ( 11 + 5q = -19 ). Then, subtract 11 from both sides, resulting in ( 5q = -30 ). Finally, divide by 5 to find ( q = -6 ).
q=-6
To simplify the expression ( p^3q + 5p^3q ), combine like terms. Both terms contain ( p^3q ), so you add the coefficients: ( 1 + 5 = 6 ). Therefore, the simplified expression is ( 6p^3q ).
3q + 2z, although it could be 2*(3q+z)
It is not an equation if it does not have an equals sign. You could simplify it but not solve it.
3q+11+8q>99
3q+11+8q>99
3q + 11 + 8q > 99 11q + 11 > 99 Subtract '11' from both sides 11q > 99 - 11 11q > 88 Divide both sides by (+)11 q > 88/11 q > 8 ('q' is greater than '8'). Verification When q = 7 (8) 11(9) = 99 > 88 (True) 1
3q = 18 q = 6
steps to solve 4(3q-q): =(4)(3q+-q) =(4)(3q)+(4)(-q) =12q-4 answer: 8q
(3)(q-11) = 3q-33
3q + 5 + 2q + 5 = 65 5q + 10 = 65 5q = 55 q = 11 Check it. 33 + 5 + 22 + 5 = 65 It checks.
If -3q + 4 = 13 then 3q = -9 : q = -3 If 6q = 13 then q = 13/6 = 2 1/6 The question contains an anomaly and is invalid.
9q2 is the square of 3q - so 3q is the result of dividing 9q2 by 3q.
24pq...I don't think there's enough info to solve this.
q=-6