We know that
If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment. (Intersecting Secant-Tangent Theorem)
so
ST²=RT*QT
RT=7 in
QT=23+7-----> 30 in
ST²=7*30-----> 210
ST=√210-----> 14.49 in
the answer is
RT=14.49 in
Answer:
A is the answer - sample 1 and sample 2 are most representative because they are the same answer.
Step-by-step explanation:
I believe the answer would b y^2 x^3•12/10
Answer:
-79.8
Step-by-step explanation:
The change can be found by subracting the ending distance from the beginning distance.
In this case, its 200.6-120.8= 79.8
It is negative because the distance in the end became lower than the beginning distance.
Hope this helps!
→ 
→ 
![\sqrt{[(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}]}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B%28x_%7B2%7D-x_%7B1%7D%29%5E%7B2%7D%2B%28y_%7B2%7D-y_%7B1%7D%29%5E%7B2%7D%5D%7D%20)
![\sqrt{[(3-(-4))^{2}+(-1-(-5))^{2}]}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B%283-%28-4%29%29%5E%7B2%7D%2B%28-1-%28-5%29%29%5E%7B2%7D%5D%7D%20)
![\sqrt{[7^{2}+4^{2}]}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B7%5E%7B2%7D%2B4%5E%7B2%7D%5D%7D%20)
![\sqrt{[49+16]}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B49%2B16%5D%7D%20)
