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3 years ago

For what reason would we use the distance formula to determine the type of quadrilateral?

Mathematics

1 answer:

Whitepunk [10]3 years ago

6 0

Answer:

Sides/diagonals are congruent

Step-by-step explanation:

If the distance formula is used to determine the type of quadrilateral, then we are interested in knowing whether the opposite sides are congruent or adjacent sides are congruent.

We can also use the length of diagonals to determine which type of quadrilateral.

For instance the square has all sides equal.

The diagonals of the rectangle are congruent.

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Prove that ΔABC and ΔEDC are similar. triangles ABC and DEC where angles A and E are right angles, AC equals 4, AB equals 3, BC

nikklg [1K]

Answer:B

Step-by-step explanation:

4E and LA are right angles; therefore, these angles are congruent since all right

angles are congruent. Is - 15 over 5 and 12 over 4shows the corresponding sides are proportional;

therefore, ABC ~ AEDC by the SSS Similarity Postulate.

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3 years ago

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Soloha48 [4]

Desert Shores is below the sea level and the difference is more than 600 feet

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3 years ago

A grocery store plans to use 75% of its area for food and beverages. If the grocery store has an area of 1,600 m², what is the a

zaharov [31]

Answer:

It's either 400 or 1,200.

Step-by-step explanation:

Other people are saying 400, I got 1,200 by multiplying 1,600 to 0.75 which gave me that answer. I do not know how other people got 400.

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3 years ago

What is the factored form of x^2y^3 -2y^3 - 2x^2 +4

Law Incorporation [45]

(-x^2+4)

x^2y^3-2y^3-2x^2+4=x^2-2x^2+4=

(-x^2+4)

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3 years ago

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