Answer:
Option D. w + x = z
Step-by-step explanation:
see the attached figure to better understand the problem
we have
----> equation A
----> equation B
equate equation A and equation B
Subtract y both sides
This equation satisfy the <em><u>Triangle Exterior Angle Theorem</u></em> (An exterior angle of a triangle is equal to the sum of the opposite interior angles)
Answer:
0.143059384
Step-by-step explanation:
=> ![\frac{\sqrt{19+6^2} }{7.2^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B19%2B6%5E2%7D%20%7D%7B7.2%5E2%7D)
=> ![\frac{\sqrt{19+36} }{51.84}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B19%2B36%7D%20%7D%7B51.84%7D)
=> ![\frac{\sqrt{55} }{51.84}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B55%7D%20%7D%7B51.84%7D)
=> ![\frac{7.416}{51.84}](https://tex.z-dn.net/?f=%5Cfrac%7B7.416%7D%7B51.84%7D)
=> 0.143059384
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>VX = 26</h2>
Step-by-step explanation:
If W is the midpoint of VX then VW = WX. Therefore we have the equation:
2x + 5 = 4x - 3 <em>subtract 5 from both sides</em>
2x = 4x - 8 <em>subtract 4x from both sides</em>
-2x = -8 <em>divide both sides by (-2)</em>
x = 4
VX = VW + WX
VX = 2x + 5 + 4x - 3 = (2x + 4x) + (5 - 3) = 6x + 2
Put the value of x = 4 to 6x + 2:
VX = 6(4) + 2 = 24 + 2 = 26
Answer:
It is most definitely D, The volume of a cone is one-third of the volume of a cylinder.