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3 years ago

Did not mean to post and now this bubble won't go away :)

Mathematics

2 answers:

bezimeni [28]3 years ago

8 0

Responding for points!

viktelen [127]3 years ago

6 0

awwww..... that's what happens if you can't restrain yourself from judgment to one person . it will imprint on itself nor the sayings .

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Given f(x) = 4x + 2, find f-1(3).

Alexxandr [17]

Answer:

1/4

Step-by-step explanation:

The inverse function tells you the input value that corresponds to the given output value of the function.

<h3>Application</h3>

You want f^-1(3), which is the value of x such that f(x) = 3.

f(x) = 3 = 4x +2

1 = 4x . . . . . . subtract 2

1/4 = x . . . . . divide by 4

The inverse function of 3 is 1/4:

$\boxed{f^{-1}(3)=\dfrac{1}{4}}$

4 0

2 years ago

Antonio had his hair cut at the salon. After he returned home, he realized he had accidentally tipped the stylist 114% of the co

stepan [7]

=>C

1.14*20.95=23.883

5 0

3 years ago

Permutations of the word "Congratulations"

Mice21 [21]

Answer:

There are 15 letters, but if the two A's must always be together, that's the same as if they're just one letter, so our "base count" is 14! ; note that this way of counting means that we also don't need to worry about compensating for "double counting" identical permutations due to transposition of those A's, because we don't "count" both transpositions. However, that counting does "double count" equivalent permutations due to having two O's, two N's, and two T's, so we do need to compensate for that. Therefore the final answer is 14!/(23)=10,897,286,400

5 0

2 years ago

Need help asap!!!!!!!!!!!

salantis [7]

Answer:

the answer should be d for tour answer

3 0

3 years ago

Express the negations of each of these statements so that all negation symbols immediately precede predicates. a) ∀x∃y∀zT(x, y,

Gnesinka [82]

Answer:

a) ∀x∃y ¬∀zT(x, y, z)

∀x∃y ∃z ¬T(x, y, z)

b) ∀x¬[∃y (P(x, y) ∨ Q(x, y))]

∀x∀y ¬ [P(x, y) ∨ Q(x, y)]

∀x∀y [¬P(x, y) ^ ¬Q(x, y)]

c) ∀x ¬∃y (P(x, y) ^ ∃zR(x, y, z))

∀x ∀y ¬(P(x, y) ^ ∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ¬∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ∀z¬R(x, y, z))

d) ∀x¬∃y (P(x, y) → Q(x, y))

∀x∀y ¬(P(x, y) → Q(x, y))

∀x∀y (¬P(x, y) ^ Q(x, y))

3 0

3 years ago

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