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iragen [17]
2 years ago
5

Given the functions, fx) = 3x² - 7 and g(x) = x³ + 1, evaluate f(g(2)).

Mathematics
1 answer:
zepelin [54]2 years ago
5 0
Answer: X squared The explanation is i already did the test for this problem
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13. All students in a class play Scrabble or Checkers or both, 36% of the students play Scrabble only:
Genrish500 [490]

Answer:

b) 16%

Step-by-step explanation:

add 36 to 48 (84) then subtract that from 100. 16% played both games

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3 years ago
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X²-12x+a=(x+b)² find a and b​
Valentin [98]

Answer:

Step-by-step explanation:

x² - 12x + a = (x + b)²

x² - 12x + a = x² + 2xb + b²

-12 = 2b and a = b²

b = -6

a = b² = 36

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3 years ago
The forumal below is used to convert a temputure in degrees celsius, C, to a tempurturein degrees fahrenheit, F.
Vika [28.1K]

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59 F

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3 years ago
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
A cell phone company has a basic monthly plan of $40 plus $0.45 for any minutes used over 700.Before receiving his statement,Joh
Dimas [21]
0.45M+40

The M stands for the amount of minutes he went over.
8 0
3 years ago
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