Answer:
Step-by-step explanation:
Reduction to normal from using lambda-reduction:
The given lambda - calculus terms is, (λf. λx. f (f x)) (λy. Y * 3) 2
For the term, (λy. Y * 3) 2, we can substitute the value to the function.
Therefore, applying beta- reduction on "(λy. Y * 3) 2" will return 2*3= 6
So the term becomes,(λf. λx. f (f x)) 6
The first term, (λf. λx. f (f x)) takes a function and an argument, and substitute the argument in the function.
Here it is given that it is possible to substitute the resulting multiplication in the result.
Therefore by applying next level beta - reduction, the term becomes f(f(f(6)) (f x)) which is in normal form.
First thing First. You must find have a common denominator. If you want to find it then you do this
3: 3 6 9 12 15 18 21
7: 7 14 21 28
Once you found a number they have in common (21) You do this next
3/7 * 3/3 = 9/21
2/3 * 7/7 = 14/21
Now the next thing you do is subtract your numerators but not the denominators
14-9=5
5/12
you cant simplified so your done! Hope this helps!!!! <span />
Answer:
B. 5.2
Step-by-step explanation:
<em><u>:-[ </u></em><em><u>y</u></em><em><u>o</u></em><em><u>u</u></em><em><u> </u></em><em><u>w</u></em><em><u>i</u></em><em><u>l</u></em><em><u>l</u></em><em><u> </u></em><em><u>k</u></em><em><u>n</u></em><em><u>o</u></em><em><u>w</u></em><em><u> </u></em><em><u>t</u></em><em><u>o</u></em><em><u> </u></em><em><u>h</u></em><em><u>a</u></em><em><u>v</u></em><em><u>e</u></em><em><u> </u></em><em><u>b</u></em><em><u>r</u></em><em><u>a</u></em><em><u>i</u></em><em><u>n</u></em><em><u>l</u></em><em><u>y</u></em><em><u>:-) </u></em>
0.11 is the least amount.