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2 years ago

What are the zeros of the quadratic function f(x) equals 6X squared +12 X -7

Mathematics

1 answer:

tamaranim1 [39]2 years ago

7 0

The zeroes of the function will be $\rm -1 \pm \dfrac{\sqrt{78}}{6}}$

<h3>What is a Quadratic Function ?</h3>

A quadratic function is given by ax² +bx+c = f(x)

The given quadratic function is

f(x) = 6x² +12x - 7

The zeroes of the function will be given by

f(0) = 0

6x² +12x - 7 = 0

The zeroes are given by

$\rm \dfrac {-b \pm \sqrt{b^2 -4ac}}{2a}$

here

a = 6

b = 12

c = -7

Therefore the zeroes of the function will be

$\rm \dfrac {-12 \pm \sqrt{12^2 -4*6*(-7)}}{2 *6}$

$\rm \dfrac {-12 \pm \sqrt{312}}{12}$

$\rm -1 \pm \dfrac{\sqrt{78}}{6}}$

Therefore these are the two zeroes of the function.

To know more about Quadratic Function

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2 years ago

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2 years ago

Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =

Rainbow [258]

Answer:

$\sin L = 0.60$

$tan\ N = 1.33$

$\cos L = 0.80$

$\sin N = 0.80$

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

$LN = 10$

First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

$100 = 36 + LM^2$

Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

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3 years ago

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