There is no solution ,<span>a+c=-10;b-c=15;a-2b+c=-5 </span>No solution System of Linear Equations entered : [1] 2a+c=-10
[2] b-c=15
[3] a-2b+c=-5
Equations Simplified or Rearranged :<span><span> [1] 2a + c = -10
</span><span> [2] - c + b = 15
</span><span> [3] a + c - 2b = -5
</span></span>Solve by Substitution :
// Solve equation [3] for the variable c
<span> [3] c = -a + 2b - 5
</span>
// Plug this in for variable c in equation [1]
<span><span> [1] 2a + (-a +2?-5) = -10
</span><span> [1] a = -5
</span></span>
// Plug this in for variable c in equation [2]
<span><span> [2] - (-? +2b-5) + b = 15
</span><span> [2] - b = 10
</span></span>
// Solve equation [2] for the variable ?
<span> [2] ? = b + 10
</span>
// Plug this in for variable ? in equation [1]
<span><span> [1] (? +10) = -5
</span><span> [1] 0 = -15 => NO solution
</span></span><span>No solution</span>
The given operation involves just swapping the two rows, so carrying out

on the matrix gives
The events are independent. By definition, it means that knowledge about one event does not help you predict the second, and this is the case: even if you knew that you rolled an even number on the first cube, would you be more or less confident about rolling a six on the second? No.
An example in which two events about rolling cubes are dependent could be something like:
Event A: You roll the first cube
Event B: The second cube returns a higher number than the first one.
In this case, knowledge on event A does change you view on event B (and vice versa): if you know that you rolled a 6 on the first cube you don't want to bet on event B, while if you know that you rolled a 1 on the first cube, you're certain that event B will happen.
Conversely, if you know that event B has happened, you are more likely to think that the first cube rolled a small number, and vice versa.
Answer:
36π cubic inches
Step-by-step explanation:
V = 4/3 πr³
radius (r): 3
= 4/3π(3)³
= 4/3π (27)
= 36π