Try -.35 I am not sure but if it’s right go ahead
You start off with 4*(cos(x+(pi/3)))
Let's place the 4 aside for the time being:
Cos(u+j) = cos(u)*cos(j) - sin(u)*sin(j)
cos(x+(pi/3)) = cos(x)*cos(pi/3) - sin(x)*sin(pi/3)
cos(x+(pi/3)) = cos(x)/2 - (sqrt(3)/2)*sin(x)
Multiplying by the 4 we put aside:
4*cos(x+(pi/3)) = 2*cos(x) - 2*sqrt(3)*sin(x)
The correct answer is B
Answer:
the time taken for the radioactive element to decay to 1 g is 304.8 s.
Step-by-step explanation:
Given;
half-life of the given Dubnium = 34 s
initial mass of the given Dubnium, m₀ = 500 grams
final mass of the element, mf = 1 g
The time taken for the radioactive element to decay to its final mass is calculated as follows;
![1 = 500 (0.5)^{\frac{t}{34}} \\\\\frac{1}{500} = (0.5)^{\frac{t}{34}}\\\\log(\frac{1}{500}) = log [(0.5)^{\frac{t}{34}}]\\\\log(\frac{1}{500}) = \frac{t}{34} log(0.5)\\\\-2.699 = \frac{t}{34} (-0.301)\\\\t = \frac{2.699 \times 34}{0.301} \\\\t = 304.8 \ s](https://tex.z-dn.net/?f=1%20%3D%20500%20%280.5%29%5E%7B%5Cfrac%7Bt%7D%7B34%7D%7D%20%5C%5C%5C%5C%5Cfrac%7B1%7D%7B500%7D%20%3D%20%20%280.5%29%5E%7B%5Cfrac%7Bt%7D%7B34%7D%7D%5C%5C%5C%5Clog%28%5Cfrac%7B1%7D%7B500%7D%29%20%3D%20log%20%5B%280.5%29%5E%7B%5Cfrac%7Bt%7D%7B34%7D%7D%5D%5C%5C%5C%5Clog%28%5Cfrac%7B1%7D%7B500%7D%29%20%20%3D%20%5Cfrac%7Bt%7D%7B34%7D%20log%280.5%29%5C%5C%5C%5C-2.699%20%3D%20%5Cfrac%7Bt%7D%7B34%7D%20%28-0.301%29%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B2.699%20%5Ctimes%2034%7D%7B0.301%7D%20%5C%5C%5C%5Ct%20%3D%20304.8%20%5C%20s)
Therefore, the time taken for the radioactive element to decay to 1 g is 304.8 s.
I think you want to describe the sentence in one word. O am answering the question based on this assumption. "Danger" is the word that best describes the situation in question. The correct option among all the options that are given in the question is the first option or option "a". I hope the answer has helped you.
Answer:
b- x=5
Step-by-step explanation:
14-5=8 so your answer would be b
