-5(11-3t) = -6t -4
distribute -5 to the parentheses
-55+15t = -6t -4
isolate t; add -6t from both sides
-55+21t = -4
add 55 from both sides
21t = 51
divide t
t = 2.4 (rounded)
Function 1 has a maximum at y = 1
Now we need to find the maximum of Function 2 by completing the square:
-x^2 + 2x - 3
= -(x^2 - 2x) - 3
= -(x - 1)^2 +1 - 3
= -(x - 1)^2 - 2
Therefor the turning point is at (1, -2) and the maximum is at y = -2
-2 < 1, therefor Function 1 has the larger maximum
The answer is b your welcome
Since LM = AM, point M must be on the perpendicular bisector of AL. Since AM = BM, BL must be perpendicular to AL. This makes ∆ALC a right triangle with hypotenuse AC twice the length of side AL. Hence ∠LAC = ∠LAB = 60°, and AL is angle bisector, median, and altitude.
ΔABC is isosceles with ∠A = 120°, and ∠B = ∠C = 30°.
Answer:
C(n, 2)a^(n-2)b^2
Step-by-step explanation:
Generally, the expansion is written in decreasing powers of "a", so the first few terms would have variable constellations that look like ...
a^n, a^(n-1)b, a^(n-2)b^2, ...
The coefficients would be (in order), C(n, k) for k increasing from 0, so the coefficient of the 3rd term would be C(n, 2).
