Answer:
Beta 17,000K, bc warmer is more blue
0.00032cm*4.02=1.2864 × 10^-3 in scientific notation.
Answer:
#include <iostream>
#include <vector>
using namespace std;
int main() {
const int NUM_GUESSES = 3;
vector<int> userGuesses(NUM_GUESSES);
int i = 0;
int uGuess = 0;
for(i = 0; i <= userGuesses.size() - 1; i++){
cin >> uGuess;
userGuesses.at(i) = uGuess;
}
cout << endl;
return 0;
}
Explanation:
First inbuilt library were imported. Then inside the main( ) function, 3 was assigned to NUM_GUESSES meaning the user is to guess 3 numbers. Next, a vector was defined with a size of NUM_GUESSES.
Then a for-loop is use to receive user guess via cin and each guess is assigned to the vector.
<span>First we can find the circumference of the whole circle with a radius of 5 feet.
circumference = 2 pi radius
circumference = (2 pi) (5 feet)
circumference = (10 pi) feet
From one high point to the other high point, the string moves through an angle of 10 degrees. Since a full circle is 360 degrees, this angle is 1/36 of a full circle.
Therefore, the arc length is 1/36 of the whole circumference.
arc length = (1/36) (circumference)
arc length = (1/36) (10 pi) feet
arc length = 0.873 feet</span>
Explanation:
We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.
Condition 1: The stars rise and set perpendicular to the horizon
The observer is at the equator
Condition 2: The stars circle the sky parallel to the horizon
The observer is at the Pole of the Earth
Condition 3: The celestial equator passes through the zenith
The observer is at the equator
Condition 4: In the course of a year, all stars are visible
The observer is at the equator
Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)
The observer is at North Pole
