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Law Incorporation [45]
2 years ago
7

The sum of 3 consecutive

Mathematics
2 answers:
Troyanec [42]2 years ago
7 0
The three numbers are 52, 53, 54.

52+53=105 105+54=159
Dennis_Churaev [7]2 years ago
6 0

Answer:

52, 53, 54

Step-by-step explanation:

To find the 3 numbers, we first use the equation x + x + 1 + x + 2 = 159.

Step 1: Combine like terms.

  • (x+x+x)+(1+2)=159
  • 3x + 3 = 159

Step 2: Subtract 3 from both sides.

  • 3x + 3 - 3 = 159 - 3
  • 3x = 156

Step 3: Divide both sides by 3.

  • 3x/3 = 156/3
  • x = 52

Step 4: Plug in 52 back into the equation.

  • (52) + (52+1) + (52+2) = 159
  • 52+53+54=159
  • 105 + 54 = 159
  • 159

Therefore, the numbers are 52, 53, and 54.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

 ~ ren ⚘

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  (d)  5a²

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The applicable rules of exponents are ...

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4 0
3 years ago
The opposite of x is y. What is the distance between x and y on the number line?
bagirrra123 [75]

Answer:

The distance between x and y on the number line is 2x

Step-by-step explanation:

we know that

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then

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4 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
20 POINTS!!! HELP PLEASE!!!!!!! BRAINLIEST
Lady_Fox [76]

Answer: (9,5) and 2

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melomori [17]
It would cost $35 because 10.50 divided by 3 = 3.5x10=35

5 0
2 years ago
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