A = 210 + 20(h - 7) <== ur equation
if she earns 410 tutoring...
410 = 210 + 20(h - 7)
410 = 210 + 20h - 140
410 = 20h + 70
410 - 70 = 20h
340 = 20h
340/20 = h
17 = h <=== she spent 17 hrs tutoring
Answer:
Use the quadratic formula
=
−
±
2
−
4
√
2
x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}
x=2a−b±b2−4ac
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
5
2
−
4
1
+
8
=
0
5x^{2}-41x+8=0
5x2−41x+8=0
=
5
a={\color{#c92786}{5}}
a=5
=
−
4
1
b={\color{#e8710a}{-41}}
b=−41
=
8
c={\color{#129eaf}{8}}
c=8
=
−
(
−
4
1
)
±
(
−
4
1
)
2
−
4
⋅
5
⋅
8
√
2
⋅
5
2
Simplify
3
Separate the equations
4
Solve
Solution
=
8
=
1
5
Answer:
5 1/3
Step-by-step explanation:
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>
