Answer:
Step-by-step explanation:
This are similar triangle
<ABD = <ACE
CE/BD = AE/AD
33/11 = 15/AD
Let AD be x
33/11 = 15/x
Cross multiply
33x = 11×15
33x = 165
x = 165/33
x = 5
AE = AD + DE
AE = 15, AD = x = 5 and DE = y
: 15 = 5 + y
y = 15 - 5
y = 10
BD/CE = AB/AC
11/33 = 9/(x+9)
Cross multiply
11(x+9) = 9(33)
11x + 99 = 297
11x = 297 - 99
11x = 198
x = 198/11
x = 18
Answer:
x > 10
Step-by-step explanation:
Add 3 to both sides.
Hello!
This is a problem about relating circle theorems to line lengths.
We can first see that both line segment MK and CM are secants within the circle that come from a common point K.
This means that the Intersecting Secant Theorem applies here.
The Intersecting Secant Theorem states that if two secants are formed from a common point outside the circle, the length of each secant multiplied by the length of its corresponding external secant are equivalent.
We can set up the following equation.





Using this value, we can find the length of line segment MK.



Hope this helps!
Answer:
No and Yes.
Step-by-step explanation:
The inequation is y ≥-x+7 and the pair is (x,y)=(-1,-1). Then:
-1≥ -(-1)+7
-1 ≥ 1+7
-1 ≥ 8, but this is not true, so (-1,-1) is not a solution of y ≥-x+7.
On the other hand, the inequation is y>(3/4)x-5 and the pair is (5,3). Then:
3>(3/4)(5)-5
3>(15/4)-5
3> -5/4 and this is true, so (5,3) is a solution of y>(3/4)x-5.