Answer:
<em>The domain of f is (-∞,4)</em>
Step-by-step explanation:
<u>Domain of a Function</u>
The domain of a function f is the set of all the values that the input variable can take so the function exists.
We are given the function
It's a rational function which denominator cannot be 0. In the denominator, there is a square root whose radicand cannot be negative, that is, 4-x must be positive or zero, but the previous restriction takes out 0 from the domain, thus:
4 - x > 0
Subtracting 4:
- x > -4
Multiplying by -1 and swapping the inequality sign:
x < 4
Thus the domain of f is (-∞,4)
We know that
A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. ( Inscribed Quadrilateral Theorem)
so
m∠B+m∠D=180°
2x+(3x-5)=180
5x=180+5
5x=185
x=185/5
x=37°
m∠A=x+5-----> 37+5------> 42°
the answer is
m∠A is 42°
Answer:
Option C is correct.
Step-by-step explanation:
y = x^2-x-3 eq(1)
y = -3x + 5 eq(2)
We can solve by substituting the value of y in eq(2) in the eq(1)
-3x+5 = x^2-x-3
x^2-x+3x-3-5=0
x^2+2x-8=0
Now factorizing the above equation
x^2+4x-2x-8=0
x(x+4)-2(x+4)=0
(x-2)(x+4)=0
(x-2)=0 and (x+4)=0
x=2 and x=-4
Now finding the value of y by placing value of x in the above eq(2)
put x =2
y = -3x + 5
y = -3(2) + 5
y = -6+5
y = -1
Now, put x = -4
y = -3x + 5
y = -3(-4) + 5
y = 12+5
y =17
so, when x=2, y =-1 and x=-4 y=17
(2,-1) and (-4,17) is the solution.
So, Option C is correct.
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:
D. 22 Cars and 8 Vans
Step-by-step explanation:
1. You multiply 22 x 150 = 3,300.
2. Then you multiply 300 x 8 = 2,400.
3. Next, add 3,300 + 2,400
4. That equals 5700.
Hopefully, that helps!
