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astra-53 [7]
2 years ago
15

Can i eat unhealthy while i’m drinking green tea for weight loss ?

Engineering
2 answers:
Zina [86]2 years ago
6 0

Answer:I don't think it's a better idea to eat unhealthy food while having green tea because it will act as a strumbling rock in your attempt of getting weight loss.

Explanation:I don't say you have to mark my ans as brainliest but if ypu think it has really helped you plz don't forget to thank me...

storchak [24]2 years ago
6 0
Weight loss treatments vary among all people. Any health improvements could help, but I don’t think this plan is meant for fast action results.
You might be interested in
Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of
rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

                     if ( values[month] > largest ){

                 largest = values[month];

return largest;

          }

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

                     if ( values[month] < smallest ){

                 smallest = values[month];

return smallest;

          }

 

int main()

{

double rainFall[NUM_MONTHS];

 for (int month = 0; month < NUM_MONTHS; month++)

  {

     cout << "Enter the rainfall (in inches) for month #";

     cout << (month + 1) << ": ";

     cin >> rainFall[month];

 

     while (rainFall[month] < 0)

     {

      cout << "Rainfall must be 0 or more.\n"

             << "Please re-enter: ";

      cin >> rainFall[month];

     }

  }

  cout << fixed << showpoint << setprecision(2) << endl;

  cout << "The total rainfall for the year is ";

  cout << getTotal(rainFall, NUM_MONTHS)

      << " inches." << endl;

   cout << "The average rainfall for the year is ";

  cout << getAverage(rainFall, NUM_MONTHS)

      << " inches." << endl;

   int subScript;

cout << "The largest amount of rainfall was ";

  cout << getLargest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl;

  cout << "The smallest amount of rainfall was ";

  cout << getSmallest(rainFall, NUM_MONTHS, subScript)

      << " inches in month ";

  cout << (subScript + 1) << "." << endl << endl;

  return 0;

}

8 0
3 years ago
Market research is a good place to start the design process and usually involves asking questions about consumers.
ElenaW [278]
what is your question?
4 0
2 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

4 0
3 years ago
A fill covering a wide area is to be placed at the surface of this profile. The fill has a total unit weight of 20 kN/m^3 and is
goblinko [34]

Answer:

hello your question lacks some information attached below is the complete question with the required information

answer : 81.63 mm

Explanation:

settlement of the surface due to compression of the clay ( new consolidated )

= 81.63 mm

attached below is a detailed solution to the given problem

8 0
3 years ago
What do u call a bad bird
statuscvo [17]

Answer:

A buzzerd

Explanation:

4 0
2 years ago
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