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ratelena [41]
3 years ago
10

The attached program (studentsGpa.cpp) uses dynamic allocation to create an array of strings. It asks the user to enter a number

and based on the entered number it allocates the array size. Then based on that number it asks the user that many times to enter student’s names. What you need to do:Add another array of doubles to store the gpa of each student as you enter themYou need to display both the student’s name and the gpa.NOTE: must use pointer notation not array subscript. Any submission that uses array subscript won’t be graded

Engineering
1 answer:
Arisa [49]3 years ago
5 0

Question: The program was not attached to your question. Find attached of the program and the answer.

Answer:

See the explanation for the answer.

Explanation:

#include <iostream>

using namespace std;

int main()

{

   cout << "How many students will you enter? ";

   int n;

   cin>>n;

   string *name = new string[n];

   double *gpa = new double[n];

   for(int i=0;i<n;i++){

       cout<<"Enter student"<<(i+1)<<"'s name: ";

       cin>>name[i];

       cout<<"Enter student"<<(i+1)<<"'s gpa: ";

       cin>>gpa[i];

   }

   cout<<"The list students"<<endl;

   cout<<"Name             GPA"<<endl;

   cout<<"----------------------"<<endl;

   for(int i=0;i<n;i++){

       cout<<name[i]<<"              "<<gpa[i]<<endl;

   }

   return 0;

}

OUTPUT : See the attached file.

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Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

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G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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Answer:

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dedylja [7]

Answer:

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Electromechanical systems convert <u>electrical energy</u> input into a <u>mechanical energy</u> output

Explanation:

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