Answer:-Sal can make 10 bracelets with 5 feet of yarn.
Explanation:-
We know that 1 foot = 12 inches
⇒1 inch=
Yarn used for 1 bracelet = 6 inches
Number of bracelets made in 5 feet
therefore, Sal can make 10 bracelets with 5 feet of yarn.
Answer:
36pie inch²
Step-by-step explanation:
Answer attached
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em> </em><em>:</em><em>)</em>
Answer:
The patient would receive 1.05mg of the drug weekly.
Step-by-step explanation:
First step: How many mcg of the drug would the patient receive daily?
The problem states that he takes three doses of 50-mcg a day. So
1 dose - 50mcg
3 doses - x mcg
x = 50*3
x = 150 mcg.
He takes 150mcg of the drug a day.
Second step: How many mcg of the drug would the patient receive weekly?
A week has 7 days. He takes 150mcg of the drug a day. So:
1 day - 150mcg
7 days - x mcg
x = 150*7
x = 1050mcg
He takes 1050mcg of the drug a week.
Final step: Conversion of 1050 mcg to mg
Each mg has 1000 mcg. How many mg are there in 1050 mcg? So
1mg - 1000 mcg
xmg - 1050mcg
1000x = 1050
x = 1.05mg
The patient would receive 1.05mg of the drug weekly.
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:
-4,2
Step-by-step explanation:
a reflection across the y axis means the y values stay the same and the x values change sign