We check with each options
'Or' represents the intersection of two graphs
'And' represents two separate graphs'
We have two separate shaded part in the given graph
So we ignore the options that has 'and' in between
LEts check first and second option
Simplify the first part and second part
multiply both sides by 2 .
x < 2 or 4x - 2 > = 26
solve 4x-2 > = 26
add 2 on both sides and then divide both sides by 4
4x >= 28
x >= 7
So solution is x<2 or x>=7 . that is the graph on number line
Lets check with second option
3x-3<3 or 2x+8>=22
add 3 on both sides
3x < 6
divide both sides by 3
so x< 2
2x+8>=22
subtract 8 on both sides
2x >= 30
divide both sides by 2
x >= 15
x<2 or x>=15 that does not satisfies the graph
So option A is correct
Step-by-step explanation:
i am expecting that total cards is 234
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;
}
Step-by-step explanation:
6:36 = 6/36 = 1/6
3:18 = 3/18 = 1/6
therefore this ratios are equivalents...
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