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3 years ago

How would i figure this out

Mathematics

1 answer:

arsen [322]3 years ago

4 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

To plot the line through the point, plot the point. Then find another point that has the given "rise" and "run". Draw the line through the two points.

The given point is (-7, -4). Locate that on the graph.

The slope is given as -2/3. This is the ratio of "rise" to "run", so it means the "rise" will be -2 for each "run" of 3. (rise/run = -2/3)

The rise is the vertical change. So, you want your second point to be 2 units below the given point. Its y-coordinate will be -4-2 = -6.

The run is the horizontal change. Your second point will be 3 units to the right of the given point, so its x-coordinate will be -7+3 = -4. Now, you can plot the point (-4, -6) and draw your graph through these two points.

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Find the difference of 238.3 and 45.098

Eddi Din [679]

Doesn’t difference mean just to subtract the two??

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3 years ago

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3 years ago

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

3 years ago

Payton read 1/6 of a book in 1/4 of an hour how much does Payton read in 1 hour

RUDIKE [14]

Answer:

4/6 of a book in an hour

Step-by-step explanation:

1/6 = 1/4

1/4 x 4/1 = 1

1/6 x 4/1 = 4/6

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3 years ago

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harina [27]

Answer:

$35

Step-by-step explanation:

Monday: $15

Tuesday: $10

15+10=35

$35 total

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2 years ago

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