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3 years ago

10

In a right triangle, the square of the length of one side is equal to the sum of the squares of the lengths of the other two sid

es. Write a program that prompts the user to enter the lengths of three sides of a triangle and then outputs a message indicating whether the triangle is a right triangle. Assume the user does not know the hypotenuse. You may NOT ask the user for the hypotenuse. Meaning, you have to check all 3 sides to verify that it is a right-angled triangle.

1 answer:

Dovator [93]3 years ago

8 0

Answer:

C++ code is explained below

Explanation:

/Remove the following line

//if not using visual studio.

#include "stdafx.h"

//Import the required header files.

#include <iostream>

using namespace std;

//Define the main function.

int main(void)

{

//Define the variables.

double side1, side2, side3;

//Prompt the user to enter the sides of the triangle.

cout << "Enter the side of the triangle:" << endl;

cout << "Enter first side of triangle:" << endl;

cin >> side1;

cout << "Enter second side of triangle:" << endl;

cin >> side2;

cout << "Enter third side of triangle:" << endl;

cin >> side3;

cout << endl;

//Check if the triangle sides are valid

if ((side1 + side2 > side3) &&

(side1 + side3 > side2) &&

(side2 + side3 > side1))

{

//Check the condition for right angle triangle

if ((side3*side3 == (side1*side1 + side2*side2)) ||

(side2*side2 == (side3*side3 + side1*side1)) ||

(side1*side1 == (side3*side3 + side2*side2)))

//Display that the triangle is right-angled triangle.

cout << "It is a right-angled triangle." << endl;

else

//Display that the triangle is not the right-//angled triangle.

cout << "It is not a right-angled triangle."

<< endl;

}

else

//Display not the valid side.

cout << "Not a valid side." << endl;

//Remove the following line if

//not using visual studio.

system("pause");

return 0;

}

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trasher [3.6K]

Answer:

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Explanation:

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2 years ago

1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

kakasveta [241]

Answer:

a) $n = 1000\,users$ , b) $\Delta t_{min} = \frac{1}{30}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$ , c) $n = 10000000\,users$ , $\Delta t_{min} = \frac{1}{3000}\,h$ , $\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$ , $\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

Explanation:

a) The total number of users that can be accomodated in the system is:

$n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )$

$n = 1000\,users$

b) The length of the side of each cell is:

$l = \sqrt{1\,km^{2}}$

$l = 1\,km$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{30}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h$

c) The total number of users that can be accomodated in the system is:

$n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )$

$n = 10000000\,users$

The length of each side of the cell is:

$l = \sqrt{100\,m^{2}}$

$l = 10\,m$

Minimum time for traversing a cell is:

$\Delta t_{min} = \frac{l}{v}$

$\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }$

$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

$\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}$

$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

8 0

3 years ago

Calculate the osmotic pressure of seawater containing 3.5 wt % NaCl at 25 °C . If reverse osmosis is applied to treat seawater,

AlladinOne [14]

Answer:

Highest osmotic pressure that membrane may experience is

' =58.638 atm

Explanation:

Suppose sea-water taken is M= 1 kg

Density of water = 1000 kg/m3

Therefore Volume of water= Mass,M/Density of water

V= 1 kg/(1000 kg/m3)

V= 10-3 m3= 1 Litre

Since mass of Nacl is 3.5 wt%,Therefore in 1 kg of water

Mass present of NaCl= m= 0.035*1000 g

m= 35 g

Since molecular weight of NaCl= 58.44 g/mol =M.W.

Thus its Number of moles of Nacl= m/M.W

nNaCl= 35g/58.44 gmol-1

= 0.5989 mol

ans since volume of solution is 1 L thus concentration of NaCl is ,C= number of moles/Volume of solution in Litres

C= 0.5989mol/ 1L

=0.5989 M

Since 1 mol NaCL disssociates to form 2 moles of ions of Na+ andCl- Thus van't hoff factor i=2

And osmotic pressure = iCRT ------------------------------(1)( Where R= 0.0821 L.atm/mol.K and T= 25oC= 298.15 K)

Putting in equation 1 ,we get = 2*(0.5989 mol/L)*(0.0821 L.atm/mol.K)*298.15 K

=29.319 atm

Now as the water gets filtered out of the membrane,the water's volume decreases and concentration C of NacL increases, thus osmotic pressure also increases.Thus, at 50% water been already filtered out, the osmotic pressure at the membrane will be maximum

Thus Volume of water left after 50% is filtered out as fresh water= 0.5 L (assuming no salt passes through semi permeable membrane)

Thus New concentration of NaCl C'= 2*C

C'=2*0.5989 M

=1.1978 M

and Since Osmotic pressure is directly proportional to concentration, Thus As concentration C doubles to C', Osmotic Pressure ' also doubles from ,

Thus,Highest osmotic pressure that membrane may experience is, '=2*

=2*29.319 atm

' =58.638 atm

3 0

3 years ago

How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?

Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = $\frac{q^2Nd^2*Xn^2}{6Eo} * f$

also note that ; Pactive ∝ Nd2 (

tD = K . $\frac{Vdd}{(Vdd - Vt )^2}$ since K = constant

Hence : Nd ∝ rt

5 0

3 years ago

If angle A is a complement of angle B, angle B and angle C are vertical angles, and the supplement of C has a measure of 140°. F

Over [174]

Answer:

50°

Explanation:

Complementary angles add up to 90°.

Supplementary angles add up to 180°.

Vertical angles are equal.

A + B = 90°

B = C

C = 180° − 140°

C = 40°

B = 40°

A = 50°

7 0

3 years ago