What “r” is equal to

E

Verdich [7]

Answer:

A. Please see the attached graphs

B. 5. The equation of the graph is y = 3·x - 3

6. The equation of the graph is y = x + 3

7. The equation of the graph is y = 4/5·x - 4

8. The equation of the graph is y = 2·x - 4

9. The graph is the line with equation y = 5·x - 31

10. The graph is the line with equation y = 5·x - 14

11. The graph is the line with equation y = -2·x + 9

12. The graph is the line with equation y = 3·x + 6

Step-by-step explanation:

A. Please see the attached graphs

B. 5. The intercepts are;

(0, -3) and (1, 0)

The slope is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the coordinate points, (0, -3) and (1, 0) we have;

m = (0 - (-3))/(1 - 0) = 3

The slope = 3

From the point and slope form, of a straight line equation, we have;

y - 0 = 3(x - 1)

The equation of the graph is therefore;

y = 3·x - 3

The y-intercept occurs at (0, -3)

The x intercept occurs where y = 0

0 = 3·x - 3

x = 3/3 = 1

The x-intercept occurs at (1, 0)

The graph of the equation, y = 3·x - 3, passes through the y and x intercepts (0, -3) and (1, 0) respectively

6. The coordinate points are;

(-3, 0) and (0, -3)

The slope is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the coordinate points, (-3, 0) and (0, -3) we have;

m = ((-3) - 0)/(0 - (-3)) = -1

The slope = 1

From the point and slope form, of a straight line equation, we have;

y - 0 = 1(x - (-3)) = x + 3

y = x + 3

The y-intercept occurs at (0, 3)

The x intercept occurs where y = 0

0 = x + 3

x = -3

The x-intercept occurs at (-3, 0)

The graph of the equation, y = x + 3, passes through the y and x intercepts (0, 3) and (-3, 0) respectively

7. The coordinate points are;

(5, 0) and (0, -4)

The slope is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the coordinate points, (5, 0) and (0, -4) we have;

m = ((-4) - 0)/(0 - 5) = 4/5

The slope = 4/5

From the point and slope form, of a straight line equation, we have;

y - 0 = 4/5×(x - 5) = -4/5·x - 4

y = 4/5·x - 4

The y-intercept occurs at (0, -4)

The x intercept occurs where y = 0

0 = -4/5·x + 4

-4 = -4/5·x

x = 5

The x-intercept occurs at (5, 0)

The graph of the equation, y = 4/5·x - 4, passes through the y and x intercept (0, 4) and (5, 0) respectively

8. The coordinate points are;

(2, 0) and (0, -4)

The slope is given by the following equation;

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the coordinate points, (2, 0) and (0, -4) we have;

m = ((-4) - 0)/(0 - 2) = 2

The slope =2

From the point and slope form, of a straight line equation, we have;

y - 0 = 2×(x - 2) = 2·x - 4

y = 2·x - 4

The y-intercept occurs at (0, -4)

The x intercept occurs where y = 0

0 = 2·x - 4

4 = 2·x

x = 2

The x-intercept occurs at (2, 0)

The graph of the equation, y = 2·x - 4 passes through the y and x intercept (0, -4) and (2, 0) respectively

C. Using the point and slope form

9. The slope m = 5 and the graph passes through the points (6, -1)

We have the point and slope form given as follows;

y - (-1) = 5·(x - 6)

y = 5·x - 30 - 1 = 5·x - 31

y = 5·x - 31

The graph is the line with equation y = 5·x - 31

10. The slope m = 5 and the graph passes through the points (2, -4)

We have the point and slope form given as follows;

y - (-4) = 5·(x - 2)

y = 5·x - 10 - 4 = 5·x - 14

y = 5·x - 14

The graph is the line with equation y = 5·x - 14

11. The slope m = -2 and the graph passes through the points (4, 1)

We have the point and slope form given as follows;

y - 1 = (-2)·(x - 4)

y = -2·x + 8 + 1 = -2·x + 9

y = -2·x + 9

The graph is the line with equation y = -2·x + 9

12. The slope m = 3 and the graph passes through the points (-3, -3)

We have the point and slope form given as follows;

y - (-3) = 3·(x - (-3))

y = 3·x + 9 - 3 = 3·x + 6

y = 3·x + 6

The graph is the line with equation y = 3·x + 6

3 0

3 years ago

I am so confused and im really struggling and this is the only question i need

icang [17]

A polynomial of degree 3 has exactly 3 solutions if we use complex numbers. If the polynomial has real coefficients and z is a complex solution, i.e., $p(z)=0$ , the conjugate is also a solution: $p(\overline{z})=0$

So, in the first case, if we know that 1-i is a solution, the conjugate 1+i must be a solution as well.

So, we know all the roots: -5, 1+i, 1-i.

When you know the solutions $x_1,\ x_2,\ldots,\ x_n$ of a polynomial, you can write the polynomial (up to multiples) as

$p=(x-x_1)(x-x_2)\ldots(x-x_n)$

So, in your case, we have

$p(x) = (x+5)(x-1-i)(x-1+i) = x^3+3x^2-8x+10$

You can check that indeed $p(0)=10$ , because the constant term is 10, so, we're ok with this

The second exercise is exactly the same: the solutions -6, i, -i identify the polynomial

$p(x)=(x+6)(x-i)(x+i)=x^3+6x^2+x+6$

In this case, P(-3)=30, so if we want P(-3)=60 we have to multiply everything by 2:

$p(x)=2x^3+12x^2+2x+12$

3 0

3 years ago

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Arisa [49]

the answer is........ B

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