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meriva
2 years ago
6

a delivery person uses a service elevator to bring boxes of books up to an office. the delivery person weighs 160 lb and each bo

x of books weighs 50 lb. the maximum capacity of the elevator is 1400 lb. how many boxes can the delivery person bring up at one time?
Mathematics
1 answer:
Semmy [17]2 years ago
4 0
He can carry around 6 boxes. How? So add one box of books and himself. You should get 210 lb. Afterwards, do 1,400 lb / 210 lb, since the max capacity of the elevator is 1,400 lb. You will get 6.666 and repeating. It will only be 6 boxes at one time. Do NOT round the decimal. The elevator will not be able to take 7 boxes PLUS himself. Otherwise the elevator will crash and fall. Hope this helped!
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kozerog [31]

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Step-by-step explanation:

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x=\dfrac{-b+\sqrt{D}}{2a} \ ; \ x=\dfrac{-b-\sqrt{D}}{2a}\\\\\\x=\dfrac{20+25.30}{2*4} \ ; \ x=\dfrac{20-25.30}{2*4}\\\\\\x=\dfrac{45.30}{8} \ ; \  x =\dfrac{-5.30}{8}\\\\

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Answer:

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7 0
2 years ago
Read 2 more answers
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.

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3 years ago
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