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ExtremeBDS [4]
4 years ago
13

A museum currently has 84 members. How many more members does the museum need to reach its goal of having 100 members?

Mathematics
2 answers:
Kazeer [188]4 years ago
4 0

Answer:

16 members

Step-by-step explanation:

100 - 84 = 16 members

pentagon [3]4 years ago
4 0

Answer:

16 more mombers

Step-by-step explanation:

if you take 100 and you subtract 84 then you will get 16. it can go the other way to if you add 16 to 84 then you will get 100 its basic math

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Find the two odd integers which when squared and added together make the number 1994
Nostrana [21]
X^2 = 1994

x= + - sqrt (1994) 

Unfortunately, there is no integer that can suit x.
3 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

From which we have;

x^3 + \dfrac{1}{x^3} is an integer.

4 0
3 years ago
What is the product of (1x-7)(2x^3-5x+3)
Nana76 [90]
You would distribute each number. Then afterwards simplify what you have into standard form. Once you have it in standard form you use the quadratic formula. The end answers are 1 and 7

8 0
3 years ago
I WILL GIVE BRAINLIEST HELP PLEASEEEE HELP PLEASEEE
Lady bird [3.3K]

Answer:

-39

Step-by-step explanation:

\left(-3\right)^3-\sqrt[3]{27}-4^3\left(\sqrt[3]{64}-\left(2\right)\left(2\right)\right)-\frac{\left(3\right)^3\left(2\right)}{6}

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-27-3-\frac{54}{6}

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Which gives us -39.

5 0
3 years ago
Jeremiah has $10 to spend at the football
Nookie1986 [14]

Answer:

if;

d=0 p=6

d=3 p=4

d=6 p=2

Step-by-step explanation:

just substitute the value into d and use algebra rules to solve!

7 0
3 years ago
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