let's recall that there are 180° in π radians, thus
![\bf \begin{array}{ccll} degrees&radians\\ \cline{1-2} 180&\pi \\ -36&x \end{array}\implies \cfrac{180}{-36}=\cfrac{\pi }{x}\implies -5=\cfrac{\pi }{x}\implies x=\cfrac{\pi }{-5}\implies x=-\cfrac{\pi }{5}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20degrees%26radians%5C%5C%20%5Ccline%7B1-2%7D%20180%26%5Cpi%20%5C%5C%20-36%26x%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B180%7D%7B-36%7D%3D%5Ccfrac%7B%5Cpi%20%7D%7Bx%7D%5Cimplies%20-5%3D%5Ccfrac%7B%5Cpi%20%7D%7Bx%7D%5Cimplies%20x%3D%5Ccfrac%7B%5Cpi%20%7D%7B-5%7D%5Cimplies%20x%3D-%5Ccfrac%7B%5Cpi%20%7D%7B5%7D)
Answer:
ME
Step-by-step explanation:
ME BC IMA GOD!
Answer:
1 and - 1
Step-by-step explanation:
The equation of a line passing through the origin is
y = mx ( m is the slope ( gradient ) )
y = x ← is in this form
with gradient m = 1
Parallel lines have equal gradients
Then the gradients of lines parallel to the given line is 1
Given a line with gradient m then the gradient of a line perpendicular to it is
= -
= -
= - 1
Then the gradients of lines perpendicular to this line is - 1
9514 1404 393
Answer:
Step-by-step explanation:
The measure of an inscribed angle (QTR) is half the measure of the arc it intercepts. The measure of an arc is the same as the measure of the central angle it intercepts. So, we have ...
∠QSR = 2×∠QTR
∠QSR = 2×39°
∠QSR = 78°
__
Sides SQ and SR are radii of circle S, so are the same length. That means triangle QRS is an isosceles triangle and the base angles SQR and SRQ are congruent. The sum of angles in a triangle is 180°, so we have ...
∠QSR + 2(∠SQR) = 180°
78° + 2(∠SQR) = 180° . . . . fill in the value we know
2(∠SQR) = 102° . . . . . . . . . subtract 78°
∠SQR = 51° . . . . . . . . . . . . .divide by 2
Answer:
40 units²
Step-by-step explanation:
Area of Rectangle: A = lw
Step 1: Define variables
<em>l</em> = 8
<em>w</em> = 5
Step 2: Substitute and evaluate
A = 8(5)
A = 40 u²