im not sure this is right
g'(x) = 6b(-5x + 1)^5 (-5)
g'(x) = -30b(-5x +1)^5
g''(x) = -30b(5)(-5x + 1)^4 (-5)
g''(x) = 750b (-5x +1)^4
g(x) = b(−5x + 1)6 − a
when
g(-x) = b(5x +1)6 - a
g'(x) = -30b(-5x +1)^5 = 0
-5x +1 = 0
x = 15
perpendicular lines have a slope that is a negative reciprocal
A) 4x-5y=5
subtract 4x
solve for y
-5y = -4x+5
divide by -5
y = 4/5 x+5 slope is 4/5 perpendicular slope is -5/4
y -y1 =m(x-x1) point slope form of a line
y-3 = -5/4 (x-5)
B) 5x+4y = 37
subtract 5x
4y =-5x +37
divide by 4
y =-5/4 x +37/4 slope is -5/4 perpendicular slope is 4/5
y -y1 =m(x-x1) point slope form of a line
y-3 = 4/5 (x-5)
C)4x+5y=5
subtract 4x
5y = -4x +5
divide by 5
y = -4/5 x +1
y =-4/5 x +1 slope is -4/5 perpendicular slope is 5/4
y -y1 =m(x-x1) point slope form of a line
y-3 = 5/4 (x-5)
D)5x-4y=8
subtract 5x
-4y = -5x+8
divide by -4
y = 5/4 x-2
y =5/4 x +-2 slope is 5/4 perpendicular slope is -4/5
y -y1 =m(x-x1) point slope form of a line
y-3 = -4/5 (x-5)
You want to find the time (x) when the distance from desitnation is 0:
d = 0,
d(x) = 1375 - 110x
0 = 1375 - 110x
110x = 1375
x = 1375/110
<u>x = 12.5 hrs</u>
9 - 6 + 4 - 8/3 ..,
geometric series a(n) = a1r^(n-1)
r = a(n+1)/a(n)
-6/9 = -2/3
4/-6 = -2/3
-8/3/4 = -2/3
so r = -2/3 and a1 = 9
Sn = a1(1-r^n)/(1-r) = 9(1-(-2/3)^n)/(1-(-2/3))
n is infinite Sn = 9/(5/3) = 27/5
Answer:
-2, -5, 2, then 5
Step-by-step explanation:
