First, we are given that the inscribed angle of arc CB which is angle D is equal to 65°. This is half of the measure of the arc which is equal to the measure of the central angle, ∠O.
m∠O = 2 (65°) = 130°
Also, the measure of the angles where the tangent lines and the radii meet are equal to 90°. The sum of the measures of the angle of a quadrilateral ACOB is equal to 360°.
m∠O + m∠C + m∠B + m∠A = 360°
Substituting the known values,
130° + 90° + 90° + m∠A = 360°
The value of m∠A is equal to 50°.
<em>Answer: 50°</em>
The answer should most likely be A. I think
keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above
![\bf y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{2}}x+5\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20y%20%3D%20%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B1%7D%7B2%7D%7Dx%2B5%5Cqquad%20%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Answer:
n=30
Step-by-step explanation:
Multiply all terms by the same value to eliminate fraction denominators
Cancel multiplied terms that are in the denominator
Re-order terms so constants are on the left - you end up with n=30
Answer:
3/4
Step-by-step explanation:
because there are 4 parts and only three are shaded so 3/4
3/4 or .75 or 75%
