Answer:
if x=-1 then its is NOT in the domain of h.
Step-by-step explanation:
Domain is the set of values for which the function is defined.
we are given the function
h(x) = x + 1 / x^2 + 2x + 1
h(x) = x+1 /x^2+x+x+1
h(x) = x+1/x(x+1)+1(x+1)
h(x) = x+1/(x+1)(x+1)
h(x) = x+1/(x+1)^2
So, the function h(x) is defined when x ≠ -1
Its is not defined when x=-1
So, if x=-1 then its is NOT in the domain of h.
Answer:
l = 16 cm
Step-by-step explanation:
Let's set the width of the rectangle as "w" and the length of the rectangle as "w/2 + 9".
The perimeter of a rectangle is calculated by adding up the value of all of its side lengths. Therefore, we can find the perimeter using 2l + 2w = P.
We can substitute 60 for P and w/2 + 9 for l.
- 2(w/2 + 9) + 2w = 60
- 2w/2 + 18 + 2w = 60
- w + 18 + 2w = 60
- 3w + 18 = 60
- 3w = 42
- w = 14
Substitute w = 14 into the equation for length l.
- l = w/2 + 9
- l = 14/2 + 9
- l = 16
The length of the rectangle is 16 cm.
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:
10/8, 20/16
Step-by-step explanation:
Volume of a triangular pyramid is v=1/3AH so the answer is 8
