Step-by-step explanation:
Quadrilateral ABCD is inscribed in a circle.
Opposite angles of a Quadrilateral are Supplementary.

9514 1404 393
Answer:
(5, 6) is (h, k)
Step-by-step explanation:
Vertex form is an instance of the transformation of parent function f(x) = x². It is vertically scaled by a factor of 'a', and translated so the vertex is point (h, k). That is, the transformed vertex is h units right and k units up from that of the parent function (0, 0).
Parent:
f(x) = x^2
Transformed:
f(x) = a(x -h)^2 +k
__
When you compare the form to your specific instance, you need to pay attention to what it is that you're comparing. As the attachment shows, ...
- a = 2
- -h = -5 ⇒ h = 5
- k = 6
Hence the vertex is (h, k) = (5, 6). The second attachment shows this on a graph.
Answer:
15 kilometers
Step-by-step explanation:
So we know:
For 4 days she runs 1500 meters each day.
For 3 days she runs 3 kilometers each day.
First off, lets convert 1500 meters into kilometers.
There are 1000 meters in a kilometer, so pluggin in 1500 meters:
1500/1000 = 1.5
So for 4 days she runs 1.5 kilometers each day.
Now, to find the total kilometers, we must find multiply the 1.5 kilometers by the 4 days, the 3 kilometers by the 3 days, then add those two together.
So lets do this:
1.5*4 = 6
So she ran 6 kilometers in the first 4 days.
Next we have:
3*3 = 9
So she ran 9 kilometers in the last 3 days.
Now finally add them together:
6 + 9 = 15
So she ran a total of 15 kilometers.
Hope this helps!
Answer:
5.64 × 10^5
Step-by-step explanation:
Answer:
Option 1: CD is a perpendicular bisector of AB
Step-by-step explanation:
Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.
Formula to find slope

Formula to Find Distance between two points

mAB ( represents , Slope of AB )
1. 
2. 
3. 
4. 
5. 
mAC = mBC , and C is common point , hence these three are collinear points making a straight line whole slope is 



Hence CD ⊥ AB
Also
From Point 4 and point 5 above , we see that
AC = CB
Hence CD bisect AB at C, also CD ⊥ AB
There fore
CD is a perpendicular bisector of AB
Therefor option 1 is true