The manager can do both but if he does 10 he will have 25 groups but if he does 5 he will have 50 groups so the simple way is 10 so he will have 25 groups
Answer:
44°
Step-by-step explanation:
cos B = 0.7193 = 44°
Answer:
Step-by-step explanation:
the problem is, there is no problem ;)
Answer:
Step-by-step explanation:
As an example, iodine-131 is a radioisotope with a half-life of 8 days. It decays by beta particle emission into xenon-131. After eight days have passed, half of the atoms of any sample of iodine-131 will have decayed, and the sample will now be 50% iodine-131 and 50% xenon-131.
50 grams to 25 grams is one half-life. 25 grams to 12.5 grams is another half-life. So, for 50 grams to decay to 12.5 grams, two half-lives, which would take 36 days total, would need to pass. This means each half-life for element X is 18 days.
1/3 ln(<em>x</em>) + ln(2) - ln(3) = 3
Recall that
, so
ln(<em>x</em> ¹ʹ³) + ln(2) - ln(3) = 3
Condense the left side by using sum and difference properties of logarithms:


Then
ln(2/3 <em>x</em> ¹ʹ³) = 3
Take the exponential of both sides; that is, write both sides as powers of the constant <em>e</em>. (I'm using exp(<em>x</em>) = <em>e</em> ˣ so I can write it all in one line.)
exp(ln(2/3 <em>x</em> ¹ʹ³)) = exp(3)
Now exp(ln(<em>x</em>)) = <em>x </em>for all <em>x</em>, so this simplifies to
2/3 <em>x</em> ¹ʹ³ = exp(3)
Now solve for <em>x</em>. Multiply both sides by 3/2 :
3/2 × 2/3 <em>x</em> ¹ʹ³ = 3/2 exp(3)
<em>x</em> ¹ʹ³ = 3/2 exp(3)
Raise both sides to the power of 3:
(<em>x</em> ¹ʹ³)³ = (3/2 exp(3))³
<em>x</em> = 3³/2³ exp(3×3)
<em>x</em> = 27/8 exp(9)
which is the same as
<em>x</em> = 27/8 <em>e</em> ⁹