Answer:
27
Step-by-step explanation:
The answer is x=27
The extraneous solution of startroot 4 x 41 endroot = x 5 will be A. -8.
<h3>What is an extraneous solution?</h3>
It should be noted that an extraneous solution simply means a root of a transformed equation which isn't part of the original equation.
✓4x + 41 = x + 5
Square both sides
4x + 41 = x² + 10x + 25
x² + 6x - 16
x(x + 8) - 2(x + 8) = 0
x + 8 = 0
x= 0 + 8 = 8
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Answer:
// C++ Program to arithmetic operationf on 2 Numbers using Recursion
// Comments are used for explanatory purpose
#include <bits/stdc++.h>
using namespace std;
// add10 recursive function to perform arithmetic operations
int add10(int m, int n)
{
return (m + product(n, 10)); //Result of m + n * 10
return 0;
}
// Main Methods Starts here
int main()
{
int m, n; // 2 Variables m and n declared as integer
cin>>m; // accept input for m
cin>>n; // accept input for n
cout << "Result : "<<add10(m,n); // Print results which is calculated by m + 10 * n
return 0;
}
Answer:
Step-by-step explanation:
1 litre = 1000 ml
water : Sucrose : Saline = 1000 : 269 : 61
This is simplest form. We cannot reduce this as 61 is a prime number and 1000; 269 will not come in 61th table or
We can say 100 ; 269 ; 61 don't have common term
Given - Taisha has a general goal is to burn the 280 calories.
she is varies by the 25 calories.
Find out the maximum and minimum of calories burn by the taisha.
To proof -
let us assume that the calories burn by the taisha be x.
as given the calories are varies by the 25 calories.
then the maximum calories equation becomes
x-25 = 280
x = 280 + 25
x = 305
the maximum calories burn by the taisha is 305 calories.
minimum calories equationbecomes
x + 25 = 280
x = 255
The minmum calories burn by the taisha is 255 calories.
Hence proved
