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3 years ago

Which function does lim(x->1) not exist?

Mathematics

1 answer:

Molodets [167]3 years ago

5 0

Answer:

Step-by-step explanation:

If the graph has a gap at the x value c, then the two-sided limit at that point will not exist.

If the graph has a vertical asymptote and one side of the asymptote goes toward infinity and the other goes toward negative infinity, then the limit does not exist.

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What is the quotient of 99 divided by 9081

Nataliya [291]

the answer is0.0101010101

8 0

3 years ago

Is the following UPC code valid? 375407370090. PLEASE SHOW OR EXPLAIN!!!

MAXImum [283]

Answer:

The UPC is not valid..

Step-by-step explanation:

The Universal Price Code is represented in the form of bars which is scanned when we purchase something.

The last digit of the UPC is the check digit.

Give UPC = 375407370090

To check whether the UPC is valid or not.

Step 1:

Add the digits at odd positions.

3 + 5 + 0 +3 + 0 + 9 = 20

Step 2:

Multiply the number obtained in step 1 by 3

20*3 = 60

Step 3:

The digits at even positions have to be added. The last digit has to be ignored as it is a check digit.

7 + 4 + 7 +7 +0 = 25

Step 4:

Add the numbers obtained in Step 2 and step 3

60 + 25 = 85

Step 5:

The number obtained in step 5 is subtracted from the next multiple of 10.

In the above case,

85 will be subtracted from 90.

90 - 85 = 5

For a valid UPC, the answer to step 5 is equal to the last digit of UPC.

In the above case, the answer is 5 which is not equal to zero.

So, the UPC is not Valid.

4 0

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