A teacher chooses four students of a class of 36 to

Solve the system of equations using elimination. SHOW WORK!!!!!!!!!!!!!

irinina [24]

q= 6

r= 7

work--

Multiply 1st column by -2:

-30q+8r=-124

5q+8r=86

Subtract 2nd column from the first column:

-35q= -210

Solve for q:

q=6

Subtract q=6 by any of the two equations, so:

-30q+8r=-124

-30*6+8r=-124

Solve for r:

r=7

8 0

3 years ago

Read 2 more answers

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

What is the y-intercept of the liner function? (y=3x-2)

inysia [295]

The y-intercept of the linear function y = 3x - 2 is -2

<h3>How to determine the y-intercept?</h3>

The function is given as

y = 3x - 2

The above function is a linear function, and the y-intercept is the point on the graph, where x = 0 i.e. the point (0, y)

As a general rule, linear functions are those functions that have constant rates or slopes

Next, we set x to 0, and calculate y to determine the value of the y-intercept

y = 3(0) - 2

Remove the bracket in the above equation

y = 3 * 0 - 2

Evaluate the product of 3 and 0 i.e. multiply 3 and 0

y = 0 - 2

Evaluate the difference of 0 and -2 i.e. subtract 0 from 2

y = -2

The above means that the value of y when x is 0 is -2

Hence, the y-intercept of the linear function y = 3x - 2 is -2