Answer:
Technician A is corect
Explanation:
Because he said the correct reason
Answer:
#include <iostream>//including iostream library to use functions such as cout and cin
using namespace std;
int main() {
int userInput = 0;
do
{
cout << "Enter a number < 100: " ;
cin >> userInput;
if (userInput < 100)//condition if number is less than 100
{
cout << "Your number < 100 is: " << userInput << endl;
}
} while (userInput > 100);//do while loop condition
return 0;
}
Explanation:
A do-while loop executes regardless in the first iteration. Once it has run through the first iteration, it checks if the condition is being met. If, the condition is TRUE, the loop begins the second iteration. If FALSE, the loop exits. In this case, the condition is the userInput. after the first iteration, lets say the userInput is 120, the condition userInput > 100 is true.Therefore, the loop will run again until eventually the number is less than hundred, lets say 25. In that case the condition would be 25 > 100, which would be false, so the loops will break.
Answer:
00111111101000000000000000000000
Explanation:
View Image
0 01111111 01000000000000000000000
The first bit is the sign bit. It's 0 for positive numbers and 1 for negative numbers.
The next 8-bits are for the exponents.
The first 0-126₁₀ (0-2⁷⁻¹) are for the negative exponent 2⁻¹-2⁻¹²⁶.
And the last 127-256₁₀ (2⁷-2⁸) are for the positive exponents 2⁰-2¹²⁶.
You have 1.25₁₀ which is 1.010₂ in binary. But IEEE wants it in scientific notation form. So its actually 1.010₂*2⁰
The exponent bit value is 127+0=127 which is 01111111 in binary.
The last 23-bits are for the mantissa, which is the fraction part of your number. 0.25₁₀ in binary is 010₂... so your mantissa will be:
010...00000000000000000000
Answer:
I think it is 10 percent because
This question is incomplete, the complete question is;
Find the magnitude of the steady-state response of the system whose system model is given by
dx(t)/dt + x(t) = f(t)
where f(t) = 2cos8t. Keep 3 significant figures
Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )
Explanation:
Given that;
dx(t)/dt + x(t) = f(t) where f(t) = 2cos8t
dx(t)/dt + x(t) = f(t)
we apply Laplace transformation on both sides
SX(s) + x(s) = f(s)
(S + 1)x(s) = f(s)
f(s) / x(s) = S + 1
x(s) / f(s) = 1 / (S + 1)
Therefore
transfer function = H(s) = x(s)/f(s) = 1/(S+1)
f(t) = 2cos8t → [ 1 / ( S + 1 ) ] → x(t) = Acos(8t - ∅ )
A = Magnitude of steady state output
S = jw
S = j8
so
A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )
A = 2/√65 = 0.2481
∅ = tan⁻¹( 1/1) = 45°
therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )