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const2013 [10]
4 years ago
15

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

Mathematics
1 answer:
Mumz [18]4 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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tankabanditka [31]

Answer: 2000

Step-by-step explanation:

Simple interest is calculated as:

(Principal × Rate × Time) / 100

We then slot the value into the formula. This would be:

720 = (P × 6 × 6)/100

720 × 100 = 36P

72000 = 36P

Principal = 72000/36

Principal = 2000

4 0
3 years ago
You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along
sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}

Then, the time (speed divided by distance) is:

t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1

To optimize this function we have to derive and equal to zero:

\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\  \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\

x

As d_b=x, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

7 0
4 years ago
Which expression shows the result of applying the distributive property to 3(1/5x − 1/7)?
elena55 [62]
<span>The expression that shows the result of applying the distributive property to 3(1/5x − 1/7) is:

3/5(x) - 3/7

I hope my answer has come to your help. God bless and have a nice day ahead! Feel free to ask more questions here in Brainly.
</span>
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3 years ago
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The answer is D
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Bob needs to wash the windows on his house. He has a 25-foot ladder and places the base of the ladder 10 feet from the wall on t
Andrei [34K]

Answer:

22.9 ft

Step-by-step explanation:

the height of the wall that reached by the ladder = √ (25²-10²)

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=√525

= 22.9 ft

7 0
3 years ago
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