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3 years ago

15

Consider this three-digit sequence: 943. Is the first digit the largest or the second digit the smallest, AND, is the third digi

t even or the first digit odd?

2 answers:

Veseljchak [2.6K]3 years ago

7 0

I think it is first number, the 9.

Jlenok [28]3 years ago

3 0

Imm not quite sure, ask someone else

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Helpppppppppppppppppppp

nata0808 [166]

Answer:

Step-by-step explanation:

x=100 (vertically opposite angle)

6 0

3 years ago

The diagram represents 6x2 – 7x + 2 with a factor of 2x – 1. A 2-column table with 2 rows. First column is labeled 2 x with entr

sukhopar [10]

Answer:

( 3x -2)

Step-by-step explanation:

6x^2 – 7x + 2

We know that the constant only has factors of 1 and 2

Since the middle term is negative we know that that we are subtracting

A negative times a negative is positive for the final term

A negative plus a negative is negative for the middle term

( -1 ) ( -2)

We have to determine how to break up 6x^2

1x * 6x

2x*3x

3x*2x

6x*1x

We are given that one factor is 2x-1

( 2x -1 ) ( -2)

That means the other factor of 6x^2 is 3x ( 2x*3x)

( 2x -1 ) ( 3x -2)

8 0

4 years ago

Read 2 more answers

Need help with this question

geniusboy [140]

Answer:

450

7,500 + 450

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

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

3 years ago

Which of the following statements is not true regarding the function y = 2"?

katovenus [111]

C.) I’m pretty sure hope I could help

5 0

3 years ago