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3 years ago

What do the following two equations represent? -x-3y=-1 -x-3y=-1

Mathematics

2 answers:

Natalija [7]3 years ago

5 0

An infinite solution, it has one line on a graph

densk [106]3 years ago

4 0

Infanate solution I think???

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Please help I am having a really bad day and crying all the time.

polet [3.4K]

Answer:

7 dollars

Step-by-step explanation:

I hope this helps and i hope you have better days and stop crying because life is way to short for all of that and i dont know you but you dont deserve to cry all the time you should be happy if you want to talk about anything im here!

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3 years ago

Sophia's building a skateboard ramp how many 2/5 ft long pieces can she cut from a 4 ft piece of wood what is the division expre

bagirrra123 [75]

Answer:

D. 4/(2/5)

Step-by-step explanation:

For example, if the wood was 8ft long, and was cut into 4 foot long segments, you would divide 8 by 4 to get 2, which is the logical answer. Use the same thinking here. In other words, divide the large piece into smaller pieces.

For this one, divide 4 (large) by 2/5 (small).

If simplification is needed you can make 10 pieces.

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3 years ago

Given 6, 8, and 11 as the three sides of a triangle, classify it as one of the following:

Katyanochek1 [597]

9514 1404 393

Answer:

C. Obtuse

Step-by-step explanation:

The "form factor" I use for this is ...

a^2 + b^2 - c^2 = 6^2 + 8^2 - 11^2 = 36 +64 -121 = -21

The value is negative, indicating an OBTUSE triangle.

<em>Additional comment</em>

In this expression, 'a' and 'b' are the two shortest sides (in no particular order) and 'c' is the longest side. The interpretation is ...

negative — obtuse

zero — right

positive — acute

If you're familiar with Pythagorean triples, you know that the 3-4-5 right triangle triple can be doubled to give 6-8-10. The long side of 11 is longer than the hypotenuse for the right triangle, so would correspond to a largest angle greater than 90°. The 6-8-11 triangle would be OBTUSE.

5 0

3 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, $x \in A \backslash (B \cap C)$ .