-3-(-2x) = -6 What’s X

Mathematics

1 answer:

vitfil [10]3 years ago

6 0

-3+2x = -6 (negative and negative changes to positive (+))

2x = -6+3

2x = -3

x = -3/2

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S KM ∥ JN? Why or why not?

bixtya [17]

Given: We have the given figure through which we can see

LK=16,

KJ=10,

LM=24,

MN=15

To Find: Whether KM || JN and the reasoning behind it.

Solution: Yes, KM || JN because $\frac{16}{10}= \frac{24}{15}$

Explanation:

For this solution, we use the concept of Similar Triangles.

Now, KM || JN if ΔLKM ~ ΔLJN (i.e., if ΔLKM is similar to ΔLJN).

Now, ∠MLK=∠NLJ

To prove similarity of the two triangles, we have to show that the sides are proportional. In other words, LK:KJ = LM:LN

$LK:KJ=LM:LN\\\\ \frac{LK}{KJ} =\frac{LM}{LN}\\\\\frac{16}{26}= \frac{24}{39}\\\\$

which is true as both sides simplify to $\frac{8}{13}$

Thus, we see that ΔLKM ~ ΔLJN (i.e., if ΔLKM is similar to ΔLJN).

Therefore, KM || JN.

To come to the reasoning, notice that

$\frac{LK}{LJ} =\frac{LM}{LN}\\\\\frac{LK}{LK+KJ} =\frac{LM}{LM+MN}\\\\\frac{LK+KJ}{LK} =\frac{LM+MN}{LM}\\\\1+\frac{KJ}{LK}=1+ \frac{MN}{LM}\\\\\frac{LK}{KJ} =\frac{LM}{MN}$

In other words, $\frac{16}{10}= \frac{24}{15}$

4 0

3 years ago

Read 2 more answers

two mountain bikers leave from the same parking lot and head in opposite directions on two different trails. the first rider goe

sveta [45]

Applying the required rule or theorem, it can be concluded that the second biker is farther from the parking lot. The distance of the bikers to the parking lot are:

i. First biker = 17.0 km

ii. Second biker = 20.22 km

The path of travel of both bikers would form a triangle. Applying the Pythagoras theorem to the path of the first biker would give his distance from the starting point. While applying the cosine rule to the path of second rider would gives his distance to the starting point.

Thus,

a. To determine the distance of the first biker from the parking lot.

Let the required distance be represented by x. Applying the Pythagoras theorem, we have:

$hyp^{2}$ = $adj 1^{2}$ + $adj 2^{2}$