Answer:this is gonna help you sm
Step-by-step explanation:
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If 
is a number that is both divisible by 4 and 5, then
4 and 5 are coprime, so we can use the Chinese remainder theorem to solve this system and find that 
is a solution to the system, where 
is any integer. Simply put, any multiple of 20 fits the bill.
Now, there are 11 numbers between 100 and 300 that are divisible by 20 (100, 120, 140, and so on). We have 
when 
, so the sum we want to compute is
Answer:
(a) ¬(p→¬q)
(b) ¬p→q
(c) ¬((p→q)→¬(q→p))
Step-by-step explanation
taking into account the truth table for the conditional connective:
<u>p | q | p→q </u>
T | T | T
T | F | F
F | T | T
F | F | T
(a) and (b) can be seen from truth tables:
for (a) <u>p∧q</u>:
<u>p | q | ¬q | p→¬q | ¬(p→¬q) | p∧q</u>
T | T | F | F | T | T
T | F | T | T | F | F
F | T | F | T | F | F
F | F | T | T | F | F
As they have the same truth table, they are equivalent.
In a similar manner, for (b) p∨q:
<u>p | q | ¬p | ¬p→q | p∨q</u>
T | T | F | T | T
T | F | F | T | T
F | T | T | T | T
F | F | T | F | F
again, the truth tables are the same.
For (c)p↔q, we have to remember that p ↔ q can be written as (p→q)∧(q→p). By replacing p with (p→q) and q with (q→p) in the answer for part (a) we can change the ∧ connector to an equivalent using ¬ and →. Doing this we get ¬((p→q)→¬(q→p))
Firstly, we need to draw triangle
we know that
O is a centroid
and centroid divides median into 2:1
so,
we have FO=4
so, we can plug it
now, we can find CF
CF=OC+FO
CF=8+4
CF=12
now, we can see triangle ACF is a right angled triangle
so, we can use pythagoras theorem
now, we can solve for x
Since, it is equilateral triangle
so,
we know that
E is a mid-point
so,
now, we can plug values
................Answer
