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3 years ago

Help plz help ......

Mathematics

2 answers:

Len [333]3 years ago

6 0

Answer:

Step-by-step explanation:

divide 30 by six to get how much glasses you can fill with 1 bottle then divide 50 by the answer as my friend daffy duck said, math is a mystery

FrozenT [24]3 years ago

3 0

Answer:2.

Step-by-step explanation:

Two bottles of juice went into each jar

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Which number line shows the solution set for |a|=6?

ivann1987 [24]

Answer:

the third one

Step-by-step explanation:

the absolute value of -6 is 6

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3 years ago

Pleqse help me do this thank you

SpyIntel [72]

Answer:

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Step-by-step explanation:

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4 years ago

angle A and angle B are vertical angles. If angle A= (5x – 9)° and angle B = (8x – 30) then find the value of x.

puteri [66]

Answer:

x=7

Step-by-step explanation:

5x-9=8x-30

8x-5x=-9+30

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2 years ago

The math team has 11 boys, 10 girls, and 1 coach. What is the ratio of coach to students? Group of answer choices 21/1 11/1 1/11

laila [671]

Answer:

1/21

Step-by-step explanation:

11 boys + 10 girls = 21 students

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1/21

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

Help plz help ......​

Help plz help ......