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Oksanka [162]
3 years ago
13

Plz answer and im giving a lot of points ; ; oh and its two questions in one

Mathematics
1 answer:
Natali [406]3 years ago
6 0
Please don't ask 2 questions in 1 question



remember that percent is parts out of 100 so x%=x/100
basically just remove the percent sign and move the decimal place 2 places to the left  to convert to fraction

the first one
20 dollars
4.75% tax
15% tip
it depends if you want to tip on the total with tax or without tax
it is better to top on the without-tax value since tax comes from you so
find 4.75% of 20 and add that to 15% of 20 then add that to 20 to get the proper value

scenarios
1. 20+(20 times 4.75%)+(20 times 15%)=23.95
2. ([(20 times 104.75%)] times 115%)=24.09
both are bellow $25 so $25 shoudl be enough to cover it




total cost of a spa treatment of 42 with 6% tax and 20% tip
so you could again do two things:
1. apply tax and give tip on the new amount
2. find the tip and add that to the tax of only 42

scenarios:
1.(42 times 106%) times 120%=$50.4
2. (42 times 20%) +(42 times 106%)=$52.92
it depens on what you tip on
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Prove the formula that:
azamat

Step-by-step explanation:

Given: [∀x(L(x) → A(x))] →

[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

To prove, we shall follow a proof by contradiction. We shall include the negation of the conclusion for

arguments. Since with just premise, deriving the conclusion is not possible, we have chosen this proof

technique.

Consider ∀x(L(x) → A(x)) ∧ ¬[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

We need to show that the above expression is unsatisfiable (False).

¬[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

∃x¬((L(x) ∧ ∃y(L(y) ∧ H(x, y))) → ∃y(A(y) ∧ H(x, y)))

∃x((L(x) ∧ ∃y(L(y) ∧ H(x, y))) ∧ ¬(∃y(A(y) ∧ H(x, y))))

E.I with respect to x,

(L(a) ∧ ∃y(L(y) ∧ H(a, y))) ∧ ¬(∃y(A(y) ∧ H(a, y))), for some a

(L(a) ∧ ∃y(L(y) ∧ H(a, y))) ∧ (∀y(¬A(y) ∧ ¬H(a, y)))

E.I with respect to y,

(L(a) ∧ (L(b) ∧ H(a, b))) ∧ (∀y(¬A(y) ∧ ¬H(a, y))), for some b

U.I with respect to y,

(L(a) ∧ (L(b) ∧ H(a, b)) ∧ (¬A(b) ∧ ¬H(a, b))), for any b

Since P ∧ Q is P, drop L(a) from the above expression.

(L(b) ∧ H(a, b)) ∧ (¬A(b) ∧ ¬H(a, b))), for any b

Apply distribution

(L(b) ∧ H(a, b) ∧ ¬A(b)) ∨ (L(b) ∧ H(a, b) ∧ ¬H(a, b))

Note: P ∧ ¬P is false. P ∧ f alse is P. Therefore, the above expression is simplified to

(L(b) ∧ H(a, b) ∧ ¬A(b))

U.I of ∀x(L(x) → A(x)) gives L(b) → A(b). The contrapositive of this is ¬A(b) → ¬L(b). Replace

¬A(b) in the above expression with ¬L(b). Thus, we get,

(L(b) ∧ H(a, b) ∧ ¬L(b)), this is again false.

This shows that our assumption that the conclusion is false is wrong. Therefore, the conclusion follows

from the premise.

15

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Answer:

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Answer:

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Step-by-step explanation:

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