<span>Northwest is 45 degrees north of west. If a car travels a distance x miles northwest, its northerly component gains x/sqrt(2) miles and its easterly component LOSES xsqrt(2)/ miles.
Resultant height is h = 4500 + 100x/sqrt(2) - 75x/sqrt(2) = [25/sqrt(2)]x + 4500
dh/dt = [25/sqrt(2)]dx/dt = [25/sqrt(2)]v </span>
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.
KD is congruent to AT as:
<h3>angles are of equal measure</h3><h3>sides KD and TA are equal</h3><h3>sides DT and KA are equal</h3>
The figure is kite as has two pairs of same sides.
<h3>They are constructed in a way of having same angle.</h3>
