Answer:
Step-by-step explanation:
For the answer to the question above, uUnder this assumption, BE is a median. Along with any median, the centroid of a triangle is always located 2/3 of the way from the vertex to the midpoint of the opposite side, so the centroid G is 2/3 of the way from E to B.
So we have GE = (2/3)BE.
Furthermore, by segment addition, BG + GE = BE.
Since BE = 9, GE = (2/3)(9) = 6.
<span>Then BG + 6 = 9, which gives BG = 3</span>
Answer:
If I entered it in correctly it would look something like this,,
Answer:
The two horiz. tang. lines here are y = -3 and y = 192.
Step-by-step explanation:
Remember that the slope of a tangent line to the graph of a function is given by the derivative of that function. Thus, we find f '(x):
f '(x) = x^2 + 6x - 16. This is the formula for the slope. We set this = to 0 and determine for which x values the tangent line is horizontal:
f '(x) = x^2 + 6x - 16 = 0. Use the quadratic formula to determine the roots here: a = 1; b = 6 and c = -16: the discriminant is b^2-4ac, or 36-4(1)(-16), which has the value 100; thus, the roots are:
-6 plus or minus √100
x = ----------------------------------- = 2 and -8.
2
Evaluating y = x^3/3+3x^2-16x+9 at x = 2 results in y = -3. So one point of tangency is (2, -3). Remembering that the tangent lines in this problem are horizontal, we need only the y-coefficient of (2, -3) to represent this first tangent line: it is y = -3.
Similarly, find the y-coeff. of the other tangent line, which is tangent to the curve at x = -8. The value of x^3/3+3x^2-16x+9 at x = -8 is 192, and so the equation of the 2nd tangent line is y=192 (the slope is zero).
A. d÷t = rt÷t; d/t= r or r= d/t
B. l÷pr = prt÷pr; l/pr= t or t= l/pr
