Answer:
The answer to your problem will be A.
Step-by-step explanation:
The easiest variable you can solve for first is "z". Knowing that opposite angles of a quadrilateral inscribed in a circle are supplementary, subtract 93 from 180 to get z.
Z should equal 87.
The next variable we can solve for is "x". We know that inscribed angles are half the measure of their intercepting arc, so we know 93 is half of (112 + x). The equation would look like this:
93= (112 + x)/2
Multiply both sides by 2
186 = 112 + x
Subtract 112 from both sides
74 = x
Now we can apply the same method we used to find "x" to find y. Set up an equation like this:
80 = (y + x)/2
Substitute the value of x in
80 = (y + 74)/2
Multiply both sides by 2
160 = y + 74
Subtract 74 from both sides
86 = y
Hope this helps!
Answer:
The system has one solution.
Both lines have the same y-intercept.
The solution is the intersection of the 2 lines.
Step-by-step explanation:
Lines are
slope = 0.5
y intercept(put x =0) =5
slope = 1
y intercept(put x =0 ) =5
Both are linear equation thus they have only one solution. Also two non parallel lines meet only at one point.
If you solve these linear equation coordinate of point of intersection of line on graph will come.
It's important that you share the complete question. What is your goal here? Double check to ensure that you have copied the entire problem correctly.
The general equation of a circle is x^2 + y^2 = r^2. Here we know that the circle passes thru two points: (-3,2) and (1,5). Given that a third point on the circle is (-7, ? ), find the y-coordinate of this third point.
Subst. the known values (of the first point) into this equation: (-3)^2 + (2)^2 = r^2. Then 9 + 4 = 13 = r^2.
Let's check this. Assuming that the equation of this specific circle is
x^2 + y^2 = r^2 = 13, the point (1,5) must satisfy it.
(1)^2 + (5)^2 = 13 is not true, unfortunately.
(1)^2 + (5)^2 = 1 + 25 = 26 (very different from 13).
Check the original problem. If it's different from that which you have shared, share the correct version and come back here for further help.
