![\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases} 343=7\cdot 7\cdot 7\\ \qquad 7^3\\ 36=6\cdot 6\\ \qquad 6^2\\ 256=4\cdot 4\cdot 4\cdot 4\\ \qquad 4^4 \end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}} \\\\\\ \sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4} \\\\\\ 7^2+6-4^3\implies 49+6-64\implies -9](https://tex.z-dn.net/?f=%5Cbf%20343%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B36%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-256%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cbegin%7Bcases%7D%0A343%3D7%5Ccdot%207%5Ccdot%207%5C%5C%0A%5Cqquad%207%5E3%5C%5C%0A36%3D6%5Ccdot%206%5C%5C%0A%5Cqquad%206%5E2%5C%5C%0A256%3D4%5Ccdot%204%5Ccdot%204%5Ccdot%204%5C%5C%0A%5Cqquad%204%5E4%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%20%287%5E3%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B%286%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-%284%5E4%29%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B%287%5E3%29%5E2%7D%2B%5Csqrt%5B2%5D%7B%286%5E2%29%5E1%7D-%5Csqrt%5B4%5D%7B%284%5E4%29%5E3%7D%5Cimplies%20%5Csqrt%5B3%5D%7B%287%5E2%29%5E3%7D%2B%5Csqrt%5B2%5D%7B%286%5E1%29%5E2%7D-%5Csqrt%5B4%5D%7B%284%5E3%29%5E4%7D%0A%5C%5C%5C%5C%5C%5C%0A7%5E2%2B6-4%5E3%5Cimplies%2049%2B6-64%5Cimplies%20-9)
to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.
Answer:
M = 1/0.000121 = 8264.5 years
Step-by-step explanation:
M = − k ∫∞₀ teᵏᵗdt
To obtain this mean life, we'll use integration by parts to integrate the function ∫ teᵏᵗdt
∫udv = uv - ∫ vdu
u = t
du/dt = 1
du = dt
∫ dv = ∫ eᵏᵗdt
v = eᵏᵗ/k
∫udv = ∫ teᵏᵗdt
uv = teᵏᵗ/k
∫ vdu = eᵏᵗ/k
∫ teᵏᵗdt = (teᵏᵗ/k) - ∫eᵏᵗ/k
But, ∫eᵏᵗ/k = (1/k) ∫eᵏᵗ = (1/k²) eᵏᵗ = eᵏᵗ/k²
∫ teᵏᵗdt = (teᵏᵗ/k) - eᵏᵗ/k²
The rest of the calculation is done on paper in the image attached to this question
Answer:
the equation should be corrected to fit the data of the problem. With the corrected equation a mass of 0.5 grams remains after 150 years
Step-by-step explanation:
for the mass y( in grams)
y=23* (1/2)^(t/45), t ≥ 0.
the initial mass is at t=0 , then
y= 23 grams → should be 16 grams
half-life from the equation = 45 years → should be 30 years
the correct equation should be
y=16*(1/2)^(t/30), t ≥ 0
then after 150 years → t= 150
y=16*(1/2)^(150/30)= 16*(1/2)^5 = 16/32 = 0.5 grams
then a mass of 0.5 grams remains after 150 years
Cos 18 = x/25
x = 25 cos 18
x = 25(0.9510) = 23.775 ~ 23.78 cm