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KonstantinChe [14]
2 years ago
9

What is the answer need help

Mathematics
1 answer:
Eva8 [605]2 years ago
8 0

Answer:

the answer is kittens

Step-by-step explanation:

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What is 7n + 3 = 3n + 27
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7n + 3 = 3n + 27

7n -3n =27-3
4n =24
n= 6.
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3 years ago
Smith cut his peanut butter
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The answer is the square root of 101.75
Assuming that the sandwich Is a square, after cutting it diagonally, it turns into 2 triangles. To look for a missing length of a triangle, you use Pythagorean’s theorem a^2+b^2=c^2
Since c is 12 and a is 6.5 you just square 12 and square 6.5
144 42.25
Subtract it
101.75
You then have to find the square root of it. You can leave it as it is as √101.75 or just solve for it and round it to 10.09
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3 years ago
Solve for the unknown. Rewrite each phrase as a multiplication sentence. Circle the scaling factor and put a box around the numb
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3 years ago
-4y^5 + 6y^3 + 8y^2 - 2y
vitfil [10]
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8 0
3 years ago
1. Consider the following hypotheses:
Andrej [43]

Answer:

See deductions below

Step-by-step explanation:

1)

a) p(y)∧q(y) for some y (Existencial instantiation to H1)

b) q(y) for some y (Simplification of a))

c) q(y) → r(y) for all y (Universal instatiation to H2)

d) r(y) for some y (Modus Ponens using b and c)

e) p(y) for some y (Simplification of a)

f) p(y)∧r(y) for some y (Conjunction of d) and e))

g) ∃x (p(x) ∧ r(x)) (Existencial generalization of f)

2)

a) ¬C(x) → ¬A(x) for all x (Universal instatiation of H1)

b) A(x) for some x (Existencial instatiation of H3)

c) ¬(¬C(x)) for some x (Modus Tollens using a and b)

d) C(x) for some x (Double negation of c)

e) A(x) → ∀y B(y) for all x (Universal instantiation of H2)

f)  ∀y B(y) (Modus ponens using b and e)

g) B(y) for all y (Universal instantiation of f)

h) B(x)∧C(x) for some x (Conjunction of g and d, selecting y=x on g)

i) ∃x (B(x) ∧ C(x)) (Existencial generalization of h)

3) We will prove that this formula leads to a contradiction.

a) ∀y (P (x, y) ↔ ¬P (y, y)) for some x (Existencial instatiation of hypothesis)

b) P (x, y) ↔ ¬P (y, y) for some x, and for all y (Universal instantiation of a)

c) P (x, x) ↔ ¬P (x, x) (Take y=x in b)

But c) is a contradiction (for example, using truth tables). Hence the formula is not satisfiable.

7 0
3 years ago
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