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bulgar [2K]
3 years ago
10

What is the slope of the line on the graph

Mathematics
2 answers:
Gekata [30.6K]3 years ago
8 0

Answer:

the slope is -1/3

Step-by-step explanation:

point 1 = (0,4)

point 2 = (6,2)

m= y2-y1/x2-x1

m = 2 - 4/ 6- 0

m = -2/6

m = -1/3

ra1l [238]3 years ago
7 0

Answer:

-3

Step-by-step explanation:

Slope = rise over run

or

Slope = (x2-x1)/(y2-y1)

We'll use the second version. Choose any 2 points on the line. We'll choose (6, 2) and (-6, 6).

Then choose which x-coordinate is x2. and the corresponding y-coordinate is y2. x2=6, so y2 = 2. Then the other two are your 1's. x1=-6, y1 = 6.

plug in the numbers to get a solution.

(6-(-6))/2-6

subtracting a negative makes it addition, so...

(6+6)/-4

12/-4

Slope = -3

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What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.
lapo4ka [179]

Consider the equation x^6+6x^3+5=0.

First, you can use the substitution t=x^3, then x^6=(x^3)^2=t^2 and equation becomes t^2+6t+5=0. This equation is quadratic, so

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Use the made substitution again:

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You have in each brackets the expression like a^3+b^3 that is equal to (a+b)(a^2-ab+b^2). Thus,

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and the equation is

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1. D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}.

2. D_2=(-1)^2-4\cdot 1=1-4=-3.

This means that quadratic trinomials don't have real roots.

Answer: x_1=-\sqrt[3]{5} , x_2=-1

If you need complex roots, then

x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2}   ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}.

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