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tatiyna
3 years ago
6

Consider the graph of the quadratic function y = 3x2 – 3x – 6. What are the solutions of the quadratic equation 0 = 3x2 − 3x − 6

?
Mathematics
2 answers:
son4ous [18]3 years ago
7 0
Y = 3x^2 - 3x - 6 {the x^2 (x squared) makes it a quadratic formula, and I'm assuming this is what you meant...}

This is derived from:
y = ax^2 + bx + c

So, by using the 'sum and product' rule:

a × c = 3 × (-6) = -18

b = -3

Now, we find the 'sum' and the 'product' of these two numbers, where b is the 'sum' and a × c is the 'product':

The two numbers are: -6 and 3

Proof:

-6 × 3 = -18 {product}

-6 + 3 = -3 {sum}

Now, since a > 1, we divide a from the results

-6/a = -6/3 = -2

3/a = 3/3 = 1

We then implement these numbers into our equation:

(x - 2) × (x + 1) = 0 {derived from 3x^2 - 3x - 6 = 0}

To find x, we make x the subject of 0:

x - 2 = 0

OR

x + 1 = 0

Therefore:

x = 2

OR

x = -1

So the x-intercepts of the quadratic formula (or solutions to equation 3x^2 - 3x -6 = 0, to put it into your words) are 2 and -1.


We can check this by substituting the values for x:

Let's start with x = 2:

y = 3(2)^2 - 3(2) - 6
= 3(4) - 6 - 6
= 12 - 6 - 6
= 0 {so when x = 2, y = 0, which is correct}

For when x = -1:

y = 3(-1)^2 - 3(-1) - 6
= 3(1) + 3 - 6
= 3 + 3 - 6
= 0 {so when x = -1, y = 0, which is correct}
Sphinxa [80]3 years ago
4 0

Answer:

X= 2 or -1 / A on edge 2020

Step-by-step explanation:

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Flauer [41]

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3 in3

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1/2 x 1/2 x 1/2=1/8

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3 years ago
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Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤
MrMuchimi
The average rate of change of a function f(x) in an interval, a < x < b is given by
\frac{f(b) - f(a)}{b - a}

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by \frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by \frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by \frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by \frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by \frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by \frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0
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A tourist drives 70 miles along a scenic highway and then takes a 7-mile walk along a hiking trail. The average velocity driving
san4es73 [151]

Answer: T=\frac{17}{x}

Step-by-step explanation:

The formula we need to use is:

time=\frac{distance}{velocity}

We know that "T" is the total time for driving and hiking and "x" is the average velocity on the hike.

Knowing that the tourist drives 70 miles along the scenic highway and walk 7-mile walk along the hiking trail, and also knowing that the average velocity driving is 7 times that while hiking, we can conclude that  the total time is:

T=\frac{70}{7x}+\frac{7}{x}

Finally, simplifying the equation, we get that  the total time for driving and hiking as a function of the average velocity on the hike, is:

T=\frac{10}{x}+\frac{7}{x}\\\\T=\frac{10+7}{x} \\\\T=\frac{17}{x}

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