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2 years ago

Tell whether the lines for each pair of equations are parallel, perpendicular, or neither

Mathematics

1 answer:

stiks02 [169]2 years ago

4 0

Answer:

C) Neither

Step-by-step explanation:

y = -1/4x + 8

-2x + 8y = 4

To compare the slopes we need to convert the second equation to the same form as the first ( that is slope/intercept form).

-2x + 8y = 4

8y = 2x + 4

Divide through by 8:-

y = 1/4 x + 1/2

The slope of the first line is -1/4 and its 1/4 for the second line.

If the slopes were the same the lines would be parallel and if they were perpendicular their product would be = -1.

The product is 1/4 * -1/4 = -1/16

So they are neither parallel nor perpendicular.

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Read 2 more answers

Solve for x in the equation x2 - 4x-9 = 29.

erastova [34]

Answer:

$x=2+\sqrt{21}\\\\x=2-\sqrt{21}$

Step-by-step explanation:

One is given the following equation;

$x^2-4x-9=29$

The problem asks one to find the roots of the equation. The roots of a quadratic equation are the (x-coordinate) of the points where the graph of the equation intersects the x-axis. In essence, the zeros of the equation, these values can be found using the quadratic formula. In order to do this, one has to ensure that one side of the equation is solved for (0) and in standard form. This can be done with inverse operations;

$x^2-4x-9=29$

$x^2-4x-38=0$

This equation is now in standard form. The standard form of a quadratic equation complies with the following format;

$ax^2+bx+c$

The quadratic formula uses the coefficients of the quadratic equation to find the zeros this equation is as follows,

$\frac{-b(+-)\sqrt{b^2-4ac}}{2a}$

Substitute the coefficients of the given equation in and solve for the roots;

$\frac{-(-4)(+-)\sqrt{(-4)^2-4(1)(-38)}}{2(1)}$

Simplify,

$\frac{-(-4)(+-)\sqrt{(-4)^2-4(1)(-38)}}{2(1)}\\\\=\frac{4(+-)\sqrt{16+152}}{2}\\\\=\frac{4(+-)\sqrt{168}}{2}\\\\=\frac{4(+-)2\sqrt{21}}{2}\\\\=2(+-)\sqrt{21}$

Therefore, the following statement can be made;

$x=2+\sqrt{21}\\\\x=2-\sqrt{21}$

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