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Maru [420]
3 years ago
6

Write the equation of a line that passes through the point (-2,7) and perpendicular to a line that passes through the points (-6

,1) and (0,4)
Mathematics
1 answer:
Alexandra [31]3 years ago
8 0

Answer:

Step-by-step explanation:

(x₁, y₁)= (-6,1)  and (x₂,y₂) = (0,4)

Slope = \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

          = \dfrac{4-1}{0-[-6]}\\\\=\dfrac{3}{0+6}\\\\=\dfrac{3}{6}\\\\=\dfrac{1}{2}

Slope of line perpendicular to it = -1/m  

                                                     = \dfrac{-1}{\dfrac{1}{2}}\\\\=-1*\dfrac{2}{1}=-2

Equation : y =mx +b

y = -2x + b

Plugin x = -2 and y =7 in the above equation

7 = -2*(-2) + b

7 = 4 + b

7-4=b

3 = b

Equation: y =-2x + 3

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What are the solutions of this quadratic equation?X2 - 10x= -34A.r=-8, -2B.r= 5 + 3iC.r=-5 + 3iD.r=-5 + 159
ololo11 [35]

The given equation is-

x^2-10x=-34

First, we move the independent term to the other side.

x^2-10x+34=0

Now, we have to use the quadratic equation to find the solutions.-

x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Where, a = 1, b = -10, and c = 34.

Replacing these values in the formula, we have.

\begin{gathered} x_{1,2}=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(1)(34)}}{2(1)} \\ x_{1,2}=\frac{10\pm\sqrt[]{100-136}}{2}=\frac{10\pm\sqrt[]{-36}}{2} \end{gathered}

But, there's no square root of -36 because it's a negative. To solve this issue, we use complex numbers that way, we would have solutions.

x_{1,2}=\frac{10\pm\sqrt[]{36}i}{2}=\frac{10\pm6i}{2}=5\pm3i<h2>Therefore, the solutions are</h2>\begin{gathered} x_1=5+3i \\ x_2=5-3i \end{gathered}The right answer is B.
5 0
1 year ago
5+(-2)+(-5)+(-3)+1+5=
zhannawk [14.2K]

Answer:

1

Step-by-step explanation:

Step 1:

5 + ( - 2  ) + ( - 5 ) + ( - 3 ) + 1 + 5          Equation

Step 2:

5 - 2 - 5 - 3 + 1 + 5     Open parenthesis

Step 3:

5 - 10 + 1 + 5       Subtract

Step 4:

- 5 + 1 + 5     Subtract

Answer:

1        Add and Subtract

Hope This Helps :)

8 0
3 years ago
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