Answer:
∠C ≅ ∠M or ∠B ≅ ∠L
Step-by-step explanation:
You are given an angle and its opposite side as being congruent. AAS requires two congruent angles and one side, so you need another set of congruent angles (one in each triangle). It does not matter which they are. The above-listed pairs are appropriate.*
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* Since the figure cannot be assumed to be drawn to scale, either of angles B or C could be declared congruent to either of angles L or M. However, it appears that angles B and L are opposite the longest side of the triangle, so it makes good sense to declare that pair congruent. The same congruence statement (ΔBCD≅ΔLMN) would result from declaring angles C and M congruent. So, either declaration will work (matches the last answer choice.)
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AAS requires two angles and a side. One side is already marked, so we do not need any more information about sides. (The second and third answer choices can be rejected as irrelevant.)
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:
<h2>
y = -4/9</h2>
Step-by-step explanation:
Given the system of equations y = 3/2 x − 6, y = −9/2 x + 21, since both expressions are functions of y, we will equate both of them to find the variable x;
3/2 x − 6 = −9/2 x + 21,
Cross multiplying;
3(2x+21) = -9(2x-6)
6x+63 = -18x+54
collecting the like terms;
6x+18x = 54-63
24x = -9
x = -9/24
x = -3/8
To get the value of y, we will substitute x = -3/8 into any of the given equation. Using the first equation;
y = 3/2x-6
y = 3/{2(-3/8)-6}
y = 3/{(-3/4-6)}
y = 3/{(-3-24)/4}
y = 3/(-27/4)
y = 3 * -4/27
y = -4/9
Hence, the value of y is -4/9
