What is the appropriate MIPS code equivalent to given c-style co int doAdd (int t0, int tl) { return t0+tl*2; } a) sll $al,$al,l add $v0,$a0,$al jr $ra doAdd: b) mul $r1,$rl,2 add $v0,$r0,$rl jr $ra doAdd: c) sll $t1,$t1,1 add $v0,$t0,$t1 jr $ra doAdd: d) doAdd: mul $s1,$sl,2 add $v0,$s0,$s1 jr $ra

Complete the given C++ function: bool checklnRange(int a, int b){ /*returns true if any of the number a,b is in the range [20, 50] , otherwise, false.*/ }

C++

I need this code to be with python 3 1. Approach: Non recursive: //include necessary header files#include <iostream>using namespace std;//main functionint main(){    int days,buy_on_this_day ,sell_on_this_day;    //get number of days as input from user    cout<<"Enter number of days: ";    cin>>days;    int stock_price[days];    for(int i=0;i<days;i++)    {   cout<<"Enter stock_price";        cin>>stock_price[i];    }    int i=0;    for(int i=0;i<days-1;i++)    {        //comparing  current price with next day price and finding the  minima        while(i<days-1 && stock_price[i+1]<=stock_price[i])        i++;         if(i==days-1)        break;        buy_on_this_day  =i++;         while(i<days && stock_price[i]>= stock_price[i-1])        i++;        sell_on_this_day =i-1;        cout<<buy_on_this_day <<" : index of the change before  we buy"<<endl;        cout<<sell_on_this_day<<"  :index…

#include <stdio.h>int main (){  unsigned int x = 0x76543210;  char *c = (char*) &x;   printf ("*c is: 0x%x\n", *c);  if (*c == 0x10)  {    printf ("Architecture is little endian. \n");  }  else  {     printf ("Architecture is big endian. \n");  }   return 0;} void printBin(int  n){ unsigned  i; for (i = 1 << 31; i > 0; i = i / 2)  (n & i) ? printf("1") : printf("0");} int main(void){ printBin(17); printf("\n"); printBin(4);}

C++ code not Java

Q2/Write program in C++ that 10 computes the following: y=1+1/x + 2/x^2 + 3/x^3 +...........+ n/x^n

C++