Answer:
<h2>1. x = 4</h2><h2>2. x = 20</h2>
Step-by-step explanation:
1.
ΔABC and ΔAJK are similar (AA). Therefore the sides are in proportion:
We have:
AC = 1 + 4 = 5
AJ = 1
AB = 1 + x
AK = 1
Substitute:
<em>subtract 1 from both sides</em>
2.
ΔVUT and ΔVMN are similar (AA). Therefore the sides are in proportion:
We hve:
VU = x + 8
VM = x
VT = 49
VN = 49 - 14 = 35
Substitute:
<em>cross multiply</em>
<em>use the distributive property a(c + b) = ab + ac</em>
<em>subtract 35x from both sides</em>
<em>divide both sides by 14</em>
1 ton is equivalent to 2000 pounds
we have to find how many pounds 3 3/8 tons are equal to
if 1 ton --> 2000 pounds
we can break down the fraction into 3 + 3/8
then 3 tons --> 2000 x 3 = 6000 pounds
3/8 ton ---> 2000 x 3/8 = 
after adding 6000 + 750 = 6750 pounds
answer is 6750 pounds
Step-by-step explanation:
The Answer for the above question is
The four integers are 1,2,5,10. And Sum of the four integers is 18 .
<u>Method : </u>
Product of four different positive integers is 100 .
<u>First</u> of all lets break the number "100"
⇒ 100 = 50*2
⇒ 100 = 25 * 2 * 2
⇒ 100 = 5 * 5 * 2 * 2
But here we get the same integers 2 and 5 and we want different positive integers . So hereby merge 5 * 2 which is 10.
Let's get to the possible combination -
After merging 5*2 = 10 we get the answer as -
⇒ 100 = 1 * 2 * 5 * 10
Here, we have 1, 2, 5, 10. these are the positive integers whose product gives 100 .
Sum of these products is -
⇒ Sum = 1 + 2 + 5 + 10
⇒ Sum = 18 .
Sum of these four Integers is 18.
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: 20%
Step-by-step explanation:
25 = 100%
25/5 = 5
100/5 = 20
That means 80% passed and 20% didnt.
