So you would set this problem up as 10 x 9 x 8 x 7, since none of the digits can be repeated. There are 5,040 4 digit passcodes possible.
Answer:
b i believe is the right answer
Step-by-step explanation:
Answer:
asymptotes: x= -4,-2 and no holes
Step-by-step explanation:
First we need to factor the denominator of the function
y = (x-1) / (x2 + 6x + 8)
x2 + 6x + 8 = 0
delta = b2 - 4ac = 36 - 32 = 4
x1 = (-6 + 2) / 2 = -2
x2 = (-6 - 2) / 2 = -4
So we have that x2 + 6x + 8 = (x+4)(x+2)
So our function is:
y = (x-1) / (x+4)(x+2)
As there is no common expressions in the numerator and denominator, there are no holes.
The asymptotes are when the denominator is zero, so:
(x+4)(x+2) = 0
x = -4 or x = -2
Answer:
58
Step-by-step explanation:
lol
use pemdas
50 + 8
58
Remark
You need to do some trig to get the final answer. There is no way around it. You can however, solve the first part with the Pythagorean Theorem.
Step One
Find AB
AB^2 = DB^2 + AD^2
AB^2 = 7.5^2 + 4.9^2
AB^2 = 56.25 + 24.01
AB^2 = 80.26
AB = sqrt(80.26)
AB = 8.9588
Step Two
Move to the triangle on your right. Find x
Tan(C)= opposite / adjacent
Tan(55.8) = AB/adjacent
Tan(55.8) = 8.9588/ x Multiply both sides by x
x * tan(55.8) = 8.9588 Divide by tan(55.8)
x = 8.9588/tan(55.8)
x = 6.09
Answer: A
