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4 years ago

I tried solving it but I get them wrong. I forgot the formula to solve them.

Mathematics

2 answers:

WARRIOR [948]4 years ago

4 0

Plain and simple. Answer is B. :-)

ratelena [41]4 years ago

4 0

You need to work out the other angles first, before working out the values of; a, b and c.
∠x = 180 - 139 [∠'s on a straight line]
thus, ∠x = 41°
∠y = ∠x [corresponding angles]
∴ ∠y = 41°

∠h = 180 - (41 + 77 + 36) [∠'s on a straight line add to 180°]
∠h = 26°

∠p = ∠h + 77° +∠y [corresponding ∠'s]
∴ ∠p = 26 + 77 + 41
∠p = 144°

∠a = 180 - (26 + 77 + 41) [∠'s on a straight line]
∠a = 180 - 144
thus, ∠a = 36°

∠c = 36 + ∠h [corresponding ∠'s]
∠c = 36 + 26
∴ ∠c = 62°

∠b = 180 - ∠c [∠'s in a straight line add to 180°]
∠b = 180 - 62
thus, ∠b = 118°

Hope you followed through. :)

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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.

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