Find the slope of a line that is perpendicular to the line 4x + 3y - 15 = 0. -3/4 4/3 3/4

1 answer:

5 0

If the equation of the line is given in a form Ax + By + C = 0 then, the slope (m) is equal to -A / B. From the given equation of the line, A = 4 and B = 3. The slope of this line is -4/3. The line perpendicular to the given has a slope which is the negative reciprocal of the slope of the given. Hence, the slope of the perpendicular line is 3/4.

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What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3?

jenyasd209 [6]

Answer:

$x=-2-\sqrt{\dfrac{11}{2}}\ \text{and}\ x=-2+\sqrt{\dfrac{11}{2}}$

Step-by-step explanation:

My favorite way to go at this is to look at a graph. It shows the vertex at (-2, -11). Since the leading coefficient is 2, this means the roots are ...

$-2\pm\sqrt{\dfrac{11}{2}}$

where the 2 in the denominator of the radical is the leading coefficient.

You can also use other clues:

the axis of symmetry is -b/(2a) = -8/(2(2)) = -2, so answer choices C and D don't work
the single change in sign in the coefficients (+ + -) tells you there is one positive real root, so answer choice B doesn't work.

The first answer choice is the only one with values symmetrical about -2 and one of them positive.

You may be expected to use the quadratic formula:

$x=\dfrac{-b\pm\sqrt{b^-4ac}}{2a}=\dfrac{-8\pm\sqrt{8^2-4(2)(-3)}}{2(2)}\\\\=\dfrac{-8}{4}\pm\dfrac{\sqrt{88}}{4}=-2\pm\sqrt{\dfrac{11}{2}}$