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const2013 [10]
3 years ago
15

What is the smallest of the whole number primitives that can represent the decimal value 333?

Mathematics
1 answer:
Mumz [18]3 years ago
6 0

Answer:

The answer to the question is

short

Step-by-step explanation:

A primitive data type are defined types of data which are recognized by a programming language. In Java programming language, for example, the eight primitive data types are the fundamental data types which store values for data manipulation . The eight primitive data types in Java includes; Boolean, char, short, int, long, float and double.

The operations of primitive data type are predefined in the Java type system and they cannot be manipulated, hence the appropriate type is to be selected for each task. All primitives have a size limit

In the case of representing the decimal value 333, we list out the whole number primitives with their capacities as follows

  • byte; Maximum value it can represent = 127
  • char; Maximum value it can represent = 2¹⁶ - 1
  • short; Maximum value it can represent = 2¹⁵ - 1
  • int; Maximum value it can represent = 2³¹ - 1
  • long; Maximum value it can represent = 2⁶³-1
  • float; Maximum value it can represent = (2-2⁻²³)·2¹²⁷
  • double; Maximum value it can represent = (2-2⁻⁵²)·2¹⁰²³
  • Boolean; Maximum value it can represent

As seen the smallest whole number primitives that can represent the decimal value 333 is short (2¹⁵ - 1)

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Given that x and y are positive integers, solve the equation x² - 4y² = 13​
zvonat [6]
<h3>Answer:  x = 7 and y = 3</h3>

=====================================================

Explanation:

Apply the difference of squares rule

x² - 4y² = 13

x² - (2y)² = 13

(x - 2y)(x + 2y) = 13

Since x and y are positive integers, this means x-2y and x+2y are both integers as well.

The value 13 is prime. Its only factors are 1 and 13

Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:

  • A) x-2y = 1 and x+2y = 13
  • B) x-2y = 13 and x+2y = 1

----------------

Let's consider case A

We have this system of equations

\begin{cases}x-2y = 1\\x+2y = 13\end{cases}

Add the equations straight down

  • x+x becomes 2x
  • -2y+2y becomes 0y = 0 which goes away
  • 1+13 becomes 14

Therefore we have 2x = 14 solve to x = 7

From here, plug this into either equation to solve for y

x-2y = 1

7 - 2y = 1

-2y = 1-7

-2y = -6

y = -6/(-2)

y = 3

You should get the same result if you used x+2y = 13

----------------

Since we've found that x = 7 and y = 3, notice how case B is not possible

Example:  x-2y = 13 becomes 7-2(3) = 13 which is false.

Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.

-----------------

Let's check those x and y values in the original equation

x² - 4y² = 13

7² - 4*(3)² = 13

49 - 4(9) = 13

49 - 36 = 13

13 = 13

The answer is confirmed.

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Read 2 more answers
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