The easiest way to prove equivalence is to draw out a truth table and then compare the values. I'm going to show a truth table using proposition logic, it's the same result as using predicate logic.
P(x) v Q(x)
P |Q || PvQ || ~Q->P <----Notice how this column matches the PvQ but if you were to
---|---||--------||---------- <----continue the truth table with ~P->Q it would not be equivalent
T T T T
T F T T
F T T T
F F F F
Let me know if you would like an example, if the truth table doesn't help.
Answer:
B) 1, 2, 2, 1
Step-by-step explanation:
In given equation:
Left side: Right Side:
2 Na 1 Na
1 S 1 S
1 H 2 H
1 Cl 1 Cl
To equal the number of Na and H atoms on both sides, HCl and NaCl will have a coefficient of 2 to balance the equation while Na₂S and H₂S will have a coefficient of 1.
1 Na₂S + 2 HCl ⇒ 2 NaCl + 1 H₂S
I think not sure but i think the discout is $5.76
Length: 2w + 59
width: w
diagonal: (2w + 59) + 2 = 2w + 61
Length² + width² = diagonal²
(2w + 59)² + (w)² = (2w + 61)²
(4w² + 118w + 3481) + w² = 4w² + 122w + 3721
5w² + 118w + 3481 = 4w² + 122w + 3721
w² + 118w + 3481 = 122w + 3721
w² - 4w + 3481 = 3721
w² - 4w - 240 = 0
a = 1, b = -4, c = -240
w = ![[-(b) +/- \sqrt{(b)^{2} - 4(a)(c) }]/2(a)](https://tex.z-dn.net/?f=%5B-%28b%29%20%2B%2F-%20%5Csqrt%7B%28b%29%5E%7B2%7D%20%20-%204%28a%29%28c%29%20%7D%5D%2F2%28a%29)
= ![[-(-4) +/- \sqrt{(-4)^{2} - 4(1)(-240) }]/2(1)](https://tex.z-dn.net/?f=%5B-%28-4%29%20%2B%2F-%20%5Csqrt%7B%28-4%29%5E%7B2%7D%20%20-%204%281%29%28-240%29%20%7D%5D%2F2%281%29)
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since width cannot be negative, disregard 1 - 2√61
w = 1 + 2√61 ≈ 16.62
Length: 2w + 59 = 2(1 + 2√61) + 59 = 2 + 4√61 + 59 = 61 + 4√61 ≈ 92.24
Answer: width = 16.62 in, length = 92.24 in
Answer:
-5/-10
Step-by-step explanation: