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Brums [2.3K]
3 years ago
13

If the sum of x and y is 90 and y is 6 more than four times x, which of the following systems of equations could be used to solv

e for x and y?
A.x+y=90
x−4y=6

B.x+y=90
4x−y=−6

C.x+y=90
6x−y=−4

D.x+y=90
4x−y=6
Mathematics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

B

Step-by-step explanation:

x + y =90

y= 4x +6

_______

x + y = 90

4x − y = -6

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Please help i need answer for test
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Answer:

He did the complete opposite of what he needed to do

Step-by-step explanation:

The error is that when you need to find x - intercept, you need to replace the value of y with zero, not the y value.

Cause otherwise, what he did here is to find the y - intercept

If you have any questions about the way I solved it, don't hesitate to ask me in the comments below :)

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Prove A-(BnC) = (A-B)U(A-C), explain with an example​
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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.

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Answer:

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marusya05 [52]

Answer:

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2 years ago
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