6 + 9x.....a common factor in both terms is 3
3(2 + 3x) <==
The first 4 is ten times as great at the other.
The first 4 is in the thousands place, which means that it is equal to 4,000. The next 4 is in the hundreds place, which makes it worth 400. If you divide the first by the second, you get how much greater it is.
4,000/400 = 10 times.
Step-by-step explanation:
Given
f(x) = 2x - 1
f^-1 (x) = ?
Now
Let y = f(x)
or y = 2x - 1
Interchanging the role of x and y
x = 2y - 1
x + 1 = 2y
y = <u>x </u><u>+</u><u> </u><u>1</u>
2
Therefore f^-1(x) = <u>x </u><u>+</u><u> </u><u>1</u>
2
Hope it will help you :)
I don't understand what you mean
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))
