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Juli2301 [7.4K]
3 years ago
8

F(x) = x2 + 2x-5 g(x) = -3/2 - 4x + 1 What is f(x) – g(x)?

Mathematics
1 answer:
NeX [460]3 years ago
6 0

Step-by-step explanation:

Start by finding (fog)(x)

To find this function, substitute x=

x−1

4

That is g(x) into f(x)

⇒(f∘g)(x)=(

x−1

4

)

2

−2(

x−1

4

)+5

=

(x−1)

2

16

−

x−1

8

+5

Now substitute x=3

⇒(f∘g)(3)=

(3−1)

2

16

−

3−1

8

+5

=

4

16

−

2

8

+5=4−4+5=5.

Hope it helps:)

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The unit rate for doing chores is $1 per hour more than the unit rate for babysitting

Step-by-step explanation:

30 / 5 = 6

28 / 4 = 7                             the person above me was right

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Prove A-(BnC) = (A-B)U(A-C), explain with an example​
NikAS [45]

Answer:

Prove set equality by showing that for any element x, x \in (A \backslash (B \cap C)) if and only if x \in ((A \backslash B) \cup (A \backslash C)).

Example:

A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace.

B = \lbrace0,\, 1 \rbrace.

C = \lbrace0,\, 2 \rbrace.

\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}.

\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}.

Step-by-step explanation:

Proof for [x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))] for any element x:

Assume that x \in (A \backslash (B \cap C)). Thus, x \in A and x \not \in (B \cap C).

Since x \not \in (B \cap C), either x \not \in B or x \not \in C (or both.)

  • If x \not \in B, then combined with x \in A, x \in (A \backslash B).
  • Similarly, if x \not \in C, then combined with x \in A, x \in (A \backslash C).

Thus, either x \in (A \backslash B) or x \in (A \backslash C) (or both.)

Therefore, x \in ((A \backslash B) \cup (A \backslash C)) as required.

Proof for [x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]:

Assume that x \in ((A \backslash B) \cup (A \backslash C)). Thus, either x \in (A \backslash B) or x \in (A \backslash C) (or both.)

  • If x \in (A \backslash B), then x \in A and x \not \in B. Notice that (x \not \in B) \implies (x \not \in (B \cap C)) since the contrapositive of that statement, (x \in (B \cap C)) \implies (x \in B), is true. Therefore, x \not \in (B \cap C) and thus x \in A \backslash (B \cap C).
  • Otherwise, if x \in A \backslash C, then x \in A and x \not \in C. Similarly, x \not \in C \! implies x \not \in (B \cap C). Therefore, x \in A \backslash (B \cap C).

Either way, x \in A \backslash (B \cap C).

Therefore, x \in ((A \backslash B) \cup (A \backslash C)) implies x \in A \backslash (B \cap C), as required.

8 0
2 years ago
If anyone can please help me
MA_775_DIABLO [31]

Answer:

1/4

Step-by-step explanation:

5 orange + 5 mango = 10

P( mango ) = mango /total = 5/10 = 1/2

Replace

5 orange + 5 mango = 10

P(orange ) = orange /total = 5/10 = 1/2

P(mange, replace, orange) = 1/2 * 1/2 = 1/4

4 0
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