See attachment the problem can be found in there

Mathematics

1 answer:

zloy xaker [14]2 years ago

7 0

Answer:

$\frac{ - 2 {x}^{2} + 19x + 3 }{3 {x} (4x^{2} - 9 )}$

Step-by-step explanation:

$\frac{5}{6 {x}^{2} + 9x} + \frac{1}{2x - 3} - \frac{2}{3x} \\ \\ = \frac{5}{3x(2 {x} + 3)} + \frac{1}{2x - 3} - \frac{2}{3x} \\ \\ = \frac{5(2x - 3) + 1 \times 3x(2x + 3)}{3x(2 {x} + 3)(2x - 3)} - \frac{2}{3x} \\ \\ = \frac{10x - 15 + 6 {x}^{2} + 9x}{3x((2 {x} ) ^{2} - {3}^{2} )} - \frac{2}{3x} \\ \\ = \frac{6 {x}^{2} + 19x - 15}{3x(4x^{2} - 9 )} - \frac{2}{3x} \\ \\ = \frac{3x(6 {x}^{2} + 19x - 15) - 2 \times 3x(4x^{2} - 9 )}{3x(4x^{2} - 9 ) \times 3x} \\ \\ = \frac{18{x}^{3} + 57 {x}^{2} - 45x - 24x^{3} + 54x }{3x(4x^{2} - 9 ) \times 3x} \\ \\ = \frac{ - 6 {x}^{3} + 57 {x}^{2} + 9x }{9 {x}^{2} (4x^{2} - 9 )} \\ \\ = \frac{ 3x(- 2 {x}^{2} + 19x + 3) }{9 {x}^{2} (4x^{2} - 9 )} \\ \\ \huge \orange{ \boxed{= \frac{ - 2 {x}^{2} + 19x + 3 }{3 {x} (4x^{2} - 9 )} }}$

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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

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