Answer:
<em> </em><em>here </em><em>9</em><em>*</em><em>1</em><em>0</em><em>^</em><em>4</em><em> </em><em>is </em><em>a </em><em>square </em><em>number </em>
Step-by-step explanation:
<em>9</em><em>*</em><em>1</em><em>0</em><em>^</em><em>4</em><em> </em><em>=</em><em> </em><em>9</em><em>0</em><em>0</em><em>0</em><em>0</em>
<em>and </em><em>it </em><em>is </em><em>a </em><em>square </em><em>of </em><em>3</em><em>0</em><em>0</em><em> </em>
Look into my attachment
in rhombus ABCD, ∠AOB = 90°
Look at ΔAOBthe sum of inner angles = 180°
∠BAO + ∠AOB + ∠ABO = 180°
2x + 1 + 90 + 2x + 5 = 180°
4x + 96 = 180
4x = 84
x = 21
Answer:
6.298 meters
Step-by-step explanation:
Here, we are told that x is the time of the dive in minutes.
We simply first need to convert 150 seconds to minutes before substitution into the regression formula;
Kindly recall that 60 seconds = 1 minute
So 150 seconds will be 150/60 = 2.5 minutes
We now proceed to substitute into;
j = 0.010 + 2.515(x)
i = 0.010 + 2.515(2.5)
j = 0.010 + 6.2875
j = 6.2975 which go 3 decimal places is 6.298 meters
Answer:
The amount of Polonium-210 left in his body after 72 days is 6.937 μg.
Step-by-step explanation:
The decay rate of Polonium-210 is the following:
(1)
Where:
N(t) is the quantity of Po-210 at time t =?
N₀ is the initial quantity of Po-210 = 10 μg
λ is the decay constant
t is the time = 72 d
The decay rate is 0.502%, hence the quantity that still remains in Alexander is 99.498%.
First, we need to find the decay constant:
(2)
Where t(1/2) is the half-life of Po-210 = 138.376 days
By entering equation (2) into (1) we have:
Therefore, the amount of Polonium-210 left in his body after 72 days is 6.937 μg.
I hope it helps you!