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3 years ago

A dog’s heart beats 6600 times in an hour. What is their pulse rate in beats per minute?

Mathematics

2 answers:

Artist 52 [7]3 years ago

7 0

Answer:

110 beats per minute.

Step-by-step explanation:

Since we know it beats 6600 beats per hour and there are 60 minutes in an hour then we have to divide 6600 by 60 which is 110.

lozanna [386]3 years ago

3 0

Hello!

We know that there are 60 minutes in an hour, so we can solve this problem by taking the hourly heart rate and dividing it by 60 to find the beats per minute:

6600/60 = 110

The dog's heart beat is 110 beats per minute.

I hope this helps you! Have a lovely day!

- Mal

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Answer: there would be 52 yellow marbles

Step-by-step explanation:

Ok we know that there is 65 marbles in total? Right? So then we’d take 65 and think, what could we divide this by and start guessing and estimating numbers. (I guessed 5, one because it’s divisible by 65, two there is 5 times more yellow marbles than red marbles.) So 5 divided by 65 is 13. But, in order to find the amount of yellow marbles, we need to subtract the whole number by the amount of red marbles. (So 65-13 =?) if you subtract that you get 52 yellow marbles. :)

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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.

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3 years ago