Answer:
116
Step-by-step explanation:
If we assume the quotient is px³+qx²+rx+s
Then the following must hold:
(px³+qx²+rx+s)(x+1) + 14 = x⁴ + ax² - 16
From this we can establish p,q,r and s and then a. Do the multiplication, and then find that:
p=1
p+q=0
q+r=a
r+s=0
s+14=-16
Combine these, and get:
s = -30
r = 30
q = -1
a = 29
Now that we have a, we can do a normal factorization:
(x⁴ + 29x² - 16) = (x³ + 2x² + 33x + 66)·(x-2) + 116
Answer: Great
Step-by-step explanation:
If secant of theta =
<span>
<span>
<span>
1.0416666667
</span>
</span>
</span>
then
theta = 16.26 Degrees
cotangent of theta = 3.4286
We've got the first number, so now we need the others.
There is 1 possibilty for the first digit, 10 for the middle ones, and 2 for the last one.
1*10*10*2=200
200 possible codes for this briefcase
Answer:
![4\sqrt{5}](https://tex.z-dn.net/?f=4%5Csqrt%7B5%7D)
Step-by-step explanation:
![\sqrt{80} = \sqrt{2^{2}*2^{2}*5}= 4\sqrt{5}](https://tex.z-dn.net/?f=%5Csqrt%7B80%7D%20%3D%20%5Csqrt%7B2%5E%7B2%7D%2A2%5E%7B2%7D%2A5%7D%3D%204%5Csqrt%7B5%7D)