Answer:
int main()
{
double pH;
int neutral;
int base;
int acid;
cout<<"Enter a pH Value";
cin>> pH;
if(pH<7.0){
neutral =0;
base=0;
acid= 1;
}
else if (pH=7.0){
neutral =1;
base=0;
acid= 0;
}
else{
neutral =0;
base=1;
acid= 0;
}
cout <<"The neutral, Base and Acid Values are: "<<neutral<<","<<base<<","<<acid<<" Respectively"<<endl;
return 0;
}
Explanation:
Using multiple if/elseif/else statement the following problem is solved with C++
Apple uses social media to promote their products and increase user-engagement. This is a very effective method of advertising.
Explanation:
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The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: <em>155 kHz.</em>
<u>Given the following data:</u>
Note: μs represents microseconds.
<u>Conversion:</u>
1 μs = 
× 
seconds
645 μs = 
× seconds
To find corresponding frequency of this sinewave, in kHz;
Mathematically, the frequency of a waveform is calculated by using the formula;
Substituting the value into the formula, we have;
Frequency = 1550.39 Hz
Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);
<u>Conversion:</u>
1 hertz = 0.001 kilohertz
1550.3876 hertz = X kilohertz
Cross-multiplying, we have;
X = 
× 
X = 155039 kHz
To 3 significant figures;
<em>Frequency = 155 kHz</em>
Find more information: brainly.com/question/23460034
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
