Question

15 points!!! GEOMETRY help please!! Questions attached!

Answer 1

3. Look at the picture.

We have the right angle triangle. We know the sum of measures of angles in triangle is equal 180°. Therefore:

$x^o+90^o+43^o=180^o\\\\x^o+133^o=180^o\ \ \ |-133^o\\\\x^o=47^o$

$Answer:\ \boxed{47^o}$

4.
Look at the picture.

Use Pythagorean theorem:

$x^2+14^2=(10+x)^2\\\\x^2+196=10^2+2\cdot10\cdot x+x^2\ \ \ \ |-x^2\\\\196=100+20x\ \ \ |-100\\\\20x=96\ \ \ |:20\\\\x=4.8$

$Used:\ (a+b)^2=a^2+2ab+b^2$

$Answer:\ \boxed{4.8\ units}$

5.
TRUE: 1; 2; 4

6.
We find a slope of the line OP:

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\O(2;\ 6)\to x_1=2;\ y_1=6\\\\P(4;\ 3)\to x_2=4;\ y_2=3\\\\m=\dfrac{3-6}{4-2}=\dfrac{-3}{2}$

We have: $OP:\ y=-\dfrac{3}{2}x+b$

Now, we must find the slope of the line perpendicular to the line OP.
We know:

$k:y=m_1x+b;\ l:y=m_2x+c\\\\k\ \perp\ l\iff m_1m_2=-1$

therefore

$-\dfrac{3}{2}m_2=-1\ \ \ |\cdot\left(-\dfrac{2}{3}\right)\\\\m_2=\dfrac{2}{3}$

So. We have the answer! :)

$Answer:\ \boxed{y=\dfrac{2}{3}x+\dfrac{1}{3}}$