Answer:
Wut tah heal? By: uncle roger
S tep-by-step explanation:
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
Look at my picture for the work
Answer: a=8 and b=2
Please let me know if this helped!!
Answer:
I think the answer is B I guessed it.
<h2>
Step-by-step explanation:</h2>
Given equations;
y₁ = 3x - 8 -------------------(i)
y₂ = 0.5x + 7 --------------------(ii)
To fill the table, substitute the values of x into equations (i) and (ii)
=> At x = 0
y₁ = 3(0) - 8 = -8
y₂ = 0.5(0) + 7 = 7
=> At x = 1
y₁ = 3(1) - 8 = -5
y₂ = 0.5(1) + 7 = 7.5
=> At x = 2
y₁ = 3(2) - 8 = -2
y₂ = 0.5(2) + 7 = 8
=> At x = 3
y₁ = 3(3) - 8 = 1
y₂ = 0.5(3) + 7 = 8.5
=> At x = 4
y₁ = 3(4) - 8 = 4
y₂ = 0.5(4) + 7 = 9
=> At x = 5
y₁ = 3(5) - 8 = 7
y₂ = 0.5(5) + 7 = 9.5
=> At x = 6
y₁ = 3(6) - 8 = 10
y₂ = 0.5(6) + 7 = 10
=> At x = 7
y₁ = 3(7) - 8 = 13
y₂ = 0.5(7) + 7 = 10.5
=> At x = 8
y₁ = 3(8) - 8 = 16
y₂ = 0.5(8) + 7 = 11
=> At x = 9
y₁ = 3(9) - 8 = 19
y₂ = 0.5(9) + 7 = 11.5
=> At x = 10
y₁ = 3(10) - 8 = 22
y₂ = 0.5(10) + 7 = 12
The complete table is attached to this response.
(ii) To find the solution of the system of equations using the table, we find the value of x for which y₁ and y₂ are the same.
As shown in the table, that value of <em>x = 6</em>. At this value of x, the values of y₁ and y₂ are both 10.