Answer:
x = -2 , y = 4
Step-by-step explanation:
given:
5x + 2y = -2 <em>*4</em>
4x + 3y = 4 <em>*5</em>
Solving simultaneously,
20x + 8y = -8 ....eq 1
(-) 20x +15y = 20 ...eq 2
.............................................
-7y = -28
y = -28 / -7
y = 4
using equation 2,
20x +15(4) = 20
20x = -40
x = -2
Answer:
310
Step-by-step explanation:
Use the Order of Operations method. (P.E.M.D.A.S.)
Our expression: 3+2⋅4+62⋅5−1
<u>Multiply 62 and 5, and 2 and 4:</u>
3+8+310-1
<u>Add and Subtract from left to right:</u>
11+310-1
311-1
310
Answer:
I've already answered :) Please give Brainliest ;)
Step-by-step explanation:
Answer:
Algorithm
Start
Int n // To represent the number of array
Input n
Int countsearch = 0
float search
Float [] numbers // To represent an array of non decreasing number
// Input array elements but first Initialise a counter element
Int count = 0, digit
Do
// Check if element to be inserted is the first element
If(count == 0) Then
Input numbers[count]
Else
lbl: Input digit
If(digit > numbers[count-1]) then
numbers[count] = digit
Else
Output "Number must be greater than the previous number"
Goto lbl
Endif
Endif
count = count + 1
While(count<n)
count = 0
// Input element to count
input search
// Begin searching and counting
Do
if(numbers [count] == search)
countsearch = countsearch+1;
End if
While (count < n)
Output count
Program to illustrate the above
// Written in C++
// Comments are used for explanatory purpose
#include<iostream>
using namespace std;
int main()
{
// Variable declaration
float [] numbers;
int n, count;
float num, searchdigit;
cout<<"Number of array elements: ";
cin>> n;
// Enter array element
for(int I = 0; I<n;I++)
{
if(I == 0)
{
cin>>numbers [0]
}
else
{
lbl: cin>>num;
if(num >= numbers [I])
{
numbers [I] = num;
}
else
{
goto lbl;
}
}
// Search for a particular number
int search;
cin>>searchdigit;
for(int I = 0; I<n; I++)
{
if(numbers[I] == searchdigit
search++
}
}
// Print result
cout<<search;
return 0;
}
Answer:
Play 3
Step-by-step explanation:
Given:
Required [Missing from the question]:
Determine the play with the least change in yard?
To do this, we ignore the sign in front of each yard before we analyze.
So, we have:
Play 1 has a change of 4
Play 2 has 6
Play 3 has 2
Play 4 has 3
<em>From the above analysis, play has the least (which is 2)</em>
