Answer:
Step-by-step explanation:
c. consistent and dependent
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>
Answer:
6.5
Step-by-step explanation:
The third choice is appropriate.
an = 5 - 3(n - 1); all integers n ≥ 1
_____
This equation follows the form for the general term of an arithmetic sequence.
an = a1 + d(n - 1)
where a1 is the first term (corresponding to n=1), and d is the common difference. From the problem statement, a1 = 5 and d = 2 - 5 = -3.
Answer:
X-8
Im not 100% sure but i think that is the answe
