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3 years ago

Please could someone help

Mathematics

1 answer:

Morgarella [4.7K]3 years ago

7 0

1. Simplify= xy^3
2. Factorise= (2x+1)(2x-1)
3.
4.
5. Estimate= 415.96
6. Highest Common Factor
70= -70,-35,-14,-10,-7,-5,-2,-1,1,2,5,7,10,14,35,70
98= -98,-49,-14,-7,-2,-1,1,2,7,14,49,98
7. Expand: 4-4x+x^2 Simplify: x^2-4x+4
8.
9. This is actually the easiest you take both the mass and volume and divide them :) which would be 500g÷25cm^3=20g/cm3 
10.

The ones I didn't answer I am not sure about and don't wanna give you any wrong answers :) Hope I helped though!

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Step-by-step explanation:

He has five, and gives sarah four. That means that you take all but one away from 5. 5-4=1.

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Solve for m 13x - 10 M N 7x - 10

lisov135 [29]

Answer:

x = 10

Step-by-step explanation:

Given:

Arc LM = 13x - 10

Arc MN = 7x - 10

Required:

Solve for x

Solution:

Recall: half a circle = 180°

Therefore,

Arc LM + Arc MN = 180°

13x - 10 + 7x - 10 = 180

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2 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