The answer to the problem is A
A line segment from a vertex to the midpoint of the opposite side is a "median". A median divides the area of the triangle in half, as it divides the base in half without changing the altitude.
AAMC is half AABC. AADC is half AAMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
ABMC is half AABC. ABMD is half ABMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
Then, AADC = 1/4 AABC = ABMC, so AADC = ABMC by the transitive property of equality.
Answer:
Given two linearly independent vectors w and z
We want -w-z
Hence apply a negative gradient to w
w=-<-5,3> =<5,-3>
So <5,-3>-<1,4> = (<5-1>,<-3-4>)
The answer is <4,-7>
Answer:
x = 5.5
Step-by-step explanation:
Given 2 secants intersecting the circle from a point outside the circle then
The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is
x(x + 14) = 6(6 + 12)
x² + 14x = 6 × 18 = 108 ( subtract 108 from both sides )
x² + 14x - 108 = 0 ← in standard form
with a = 1, b = 14 and c = - 108
Using the quadratic formula to solve for x
x = ( - b ±
) / 2a
= ( - 14 ±
) / 2
= ( - 14 ±
) / 2
= ( - 14 ±
) / 2
x =
or x = 
x = - 19.5 or x = 5.5 ( to 1 dec. place )
However x > 0 ⇒ x = 5.5