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3 years ago

If two lines are parallel, then corresponding angles are congruent. i

Mathematics

1 answer:

Alchen [17]3 years ago

6 0

Answer: Choice B)

If corresponding angles are not congruent, then the lines are not parallel.

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Explanation:

The given conditional is in the form "If P, then Q"

where,

P = two lines are parallel
Q = corresponding angles are congruent

The P and Q are placeholders for those definitions shown above. So we'll simply do a "search and replace" to replace P and Q with those phrases. We do this shortening so that we can do logical operations.

Specifically, we'll do two things:

Swap the positions of P and Q
Place a "not" before each letter after the swap

We go from the format "if P, then Q" to "if not Q, then not P". Those two formats are equivalent in truth value. This is known as the contrapositive.

In terms of the problem at hand, the phrase "if not Q, then not P" becomes "If corresponding angles are not congruent, then the lines are not parallel"

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In short, we swap each part of the original "if ..., then ..." statement, and stick "not"s in there to negate each piece.

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The three angle bisectors of ΔABC intersect at point P. Point P is the

lisabon 2012 [21]

<h3>Answer: C) incenter</h3>

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Explanation:

If you were to intersect the angle bisectors (at least two of them), then you would locate the incenter. The incenter is the center of the incircle which is a circle where it is as large as possible, but does not spill over and outside the triangle. Therefore this circle fits snugly inside the triangle.

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extra notes:

* The centroid is found by intersecting at least two median lines

* The circumcenter is found by intersecting at least two perpendicular bisector lines

* The orthocenter is found by intersecting at least two altitude lines

* The incenter is always inside the triangle; hence the "in" as part of the name. The centroid shares this property as well because the medians are completely contained within any triangle. The other two centers aren't always guaranteed to be inside the triangle.

* The red lines cut each angle of the triangle into two equal or congruent pieces.

5 0

3 years ago

Use the distributive property to factor the expression.<br><br> 15x + 6

Marianna [84]

Take 3 out of both of them
3(5x+2)

5 0

3 years ago

If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$