I didn't get the same answers.
In this problem, 25 is your constant (the number of baby hats you already started out with).
26 will be affected by your variable, d, the number of days. With each day that passes, 26 more hats will be knit, so the expression 26d can be used.
Set h equal to the constant + hats made per day to create an expression that you can use to solve for h:
h = 26d + 25
Now, just plug the numbers in for d to get h:
When d = 2, h = 26d + 25 = 26(2) + 25 = 77
When d = 4, h = 26d + 25 = 26(4) + 25 = 129
When d = 7, h = 26d + 25 = 26(7) + 25 = 207
When d = 9, h = 26d + 25 = 26(9) + 25 = 259
Your answers should be:
77
129
207
259
Answer:
Yes
Step-by-step explanation:
A square plus b square equal c square
Answer:
I think its 20 to 60 or something like that....
Step-by-step explanation:
Answer:
Step-by-step explanation:
The Universal Set, n(U)=2092
Let the number who take all three subjects, 
Note that in the Venn Diagram, we have subtracted from each of the intersection of two sets.
The next step is to determine the number of students who study only each of the courses.
![n(S\:only)=1232-[103-x+x+23-x]=1106+x\\n(F\: only)=879-[103-x+x+14-x]=762+x\\n(R\:only)=114-[23-x+x+14-x]=77+x](https://tex.z-dn.net/?f=n%28S%5C%3Aonly%29%3D1232-%5B103-x%2Bx%2B23-x%5D%3D1106%2Bx%5C%5Cn%28F%5C%3A%20only%29%3D879-%5B103-x%2Bx%2B14-x%5D%3D762%2Bx%5C%5Cn%28R%5C%3Aonly%29%3D114-%5B23-x%2Bx%2B14-x%5D%3D77%2Bx)
These values are substituted in the second Venn diagram
Adding up all the values
2092=[1106+x]+[103-x]+x+[23-x]+[762+x]+[14-x]+[77+x]
2092=2085+x
x=2092-2085
x=7
The number of students who have taken courses in all three subjects,
Simplify 2.5 + 13.5 to 16
10 × 16
Simplify
160
