Answer:
1) 
2) ![\sqrt[3]{y^5}=y^{\frac{5}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D%3Dy%5E%7B%5Cfrac%7B5%7D%7B3%7D)
3) ![\sqrt[5]{a^{12}}=a^{\frac{12}{5} }](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D%3Da%5E%7B%5Cfrac%7B12%7D%7B5%7D%20%7D)
4) ![\sqrt[4]{z^{9}}=z^\frac{9}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D%3Dz%5E%5Cfrac%7B9%7D%7B4%7D)
Step-by-step explanation:
1) 
We know that 
So, 
2) ![\sqrt[3]{y^5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D)
We know that ![\sqrt[3]{x}=x^{\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B3%7D)
So, ![\sqrt[3]{y^5}=y^{\frac{5}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D%3Dy%5E%7B%5Cfrac%7B5%7D%7B3%7D)
3) ![\sqrt[5]{a^{12}}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D)
We know that ![\sqrt[5]{x}=x^{\frac{1}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B5%7D)
So, ![\sqrt[5]{a^{12}}=a^{\frac{12}{5} }](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D%3Da%5E%7B%5Cfrac%7B12%7D%7B5%7D%20%7D)
4) ![\sqrt[4]{z^{9}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D)
We know that ![\sqrt[4]{x}=x^{\frac{1}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B4%7D)
So, ![\sqrt[4]{z^{9}}=z^\frac{9}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D%3Dz%5E%5Cfrac%7B9%7D%7B4%7D)
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))
Answer:
6 lbs and 1 oz
Step-by-step explanation:
Answer:
answer is A
Step-by-step explanation:
All of these liquids were the same temperature when Dale took them out of the refrigerator. They all spent the same amount of time sitting on the counter, but at the end of that time, each one was a different temperature. This means that different liquids change temperatures at different rates.
Answer:
4
Step-by-step explanation:
3 x 4 = 12