For ABC:

The interior angle of C is 180 - 142 = 38
The three interior angles add up to 180°
(2x - 15) + (x - 5) + 38 = 180
3x - 20 + 38 = 180
3x + 18 = 180
3x = 162
x = 54

The measure of angle ABC = x - 5 = 54 - 5 = 49

For JKL:
The interior angle of L is 180 - 100 = 80°
The 3 interior angles add up to 180°
(2x + 27) + (2x - 11) + 80 = 180
4x + 16 + 80 = 180
4x + 96 = 180
4x = 84
x = 21

The measure of angle JKL is 2x - 11 = 2(21) - 11 = 42 - 11 = 31°

7 0

3 years ago

How to solve this?<br>\int \frac { 4 - 3 x ^ { 2 } } { ( 3 x ^ { 2 } + 4 ) ^ { 2 } } d x

ivanzaharov [21]

$\Large \mathbb{SOLUTION:}$

$\begin{array}{l} \displaystyle \int \dfrac{4 - 3x^2}{(3x^2 + 4)^2} dx \\ \\ = \displaystyle \int \dfrac{4 - 3x^2}{x^2\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ = \displaystyle \int \dfrac{\dfrac{4}{x^2} - 3}{\left(3x + \dfrac{4}{x}\right)^2} dx \\ \\ \text{Let }u = 3x + \dfrac{4}{x} \implies du = \left(3 - \dfrac{4}{x^2}\right)\ dx \\ \\ \text{So the integral becomes} \\ \\ = \displaystyle -\int \dfrac{du}{u^2} \\ \\ = -\dfrac{u^{-2 + 1}}{-2 + 1} + C \\ \\ = \dfrac{1}{u} + C \\ \\ = \dfrac{1}{3x + \dfrac{4}{x}} + C \\ \\ = \boxed{\dfrac{x}{3x^2 + 4} + C}\end{array}$

5 0

3 years ago