The linear equation for the remaining Federal Helium Reserves, r(t) = 20 -1.9t
There is 20 billion cubic feet of Helium in 2010. Helium is depleted by about 1.9 billion cubic feet each year.
Let 'r' denotes the remaining Federal Helium Reserves in billions of cubic feet.
Let 't' denotes the number of years since 2010.
We have to find a linear equation to represent the depletion of Helium through years since 2010. A linear equation will have two variables, one dependent and one independent variable. Here, let 'r' be the dependent variable and 't' be the independent variable.
Then the amount of Helium depleted in 't' years = 1.9t
So, the remaining Helium in the Federal Helium Reserve since 2010 = 20 - 1.9t.
Thus, r(t) = 20 - 1.9t
This is a linear equation because it depends on only one variable, 't'.
Learn more about linear equations at brainly.com/question/15602982
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X=number of females
47-x=number of males
# of males is 2*(number of females)-7:
47-x= 2x-7
add x to both sides:
47=3x-7
add 7 to both sides:
54=3x
divide both sides by 3:
x= 18 females
47-18= 29 males
Answer:
Total money left with Jamie = £13
Step-by-step explanation:
Jamie received pocket money = £26
He spent on sweets = 10% of total pocket money
Expense on magazines = 25% of the total money
Expense on games = 15% of the pocket money
Total expenditure on sweets, magazines and games = 
= 
Or 50% of the pocket money he received
Therefore, money left with Jamie = 50% of total money he received as pocket money
= 
= £13
It's the difference between the upper quartile and the lower quartile
Answer:
![\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B95%5E2%7D%7D%7B17%5Ccdot95%5E4%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B9%5C%2C025%7D%7D%7B1%5C%2C384%5C%2C660%5C%2C625%7D)
Step-by-step explanation:
The applicable rules of exponents are ...
(ab)^c = (a^c)(b^c)
(a^b)/(a^c) = a^(b-c)
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![\dfrac{190^3}{68^2}\times\dfrac{34}{95^{\frac{19}{3}}}=\dfrac{(2\cdot 95)^3}{(2\cdot 34)^2}\cdot\dfrac{34}{95^6\cdot 95^{\frac{1}{3}}}=2^{3-2}95^{3-6-\frac{1}{3}}34^{1-2}\\\\=2\cdot 95^{-3\frac{1}{3}}\cdot 34^{-1}=2\cdot 95^{-4+\frac{2}{3}}\cdot 34^{-1}\\\\=\dfrac{2\sqrt[3]{95^2}}{95^4\cdot 34}=\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}\\\\=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}](https://tex.z-dn.net/?f=%5Cdfrac%7B190%5E3%7D%7B68%5E2%7D%5Ctimes%5Cdfrac%7B34%7D%7B95%5E%7B%5Cfrac%7B19%7D%7B3%7D%7D%7D%3D%5Cdfrac%7B%282%5Ccdot%2095%29%5E3%7D%7B%282%5Ccdot%2034%29%5E2%7D%5Ccdot%5Cdfrac%7B34%7D%7B95%5E6%5Ccdot%2095%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%3D2%5E%7B3-2%7D95%5E%7B3-6-%5Cfrac%7B1%7D%7B3%7D%7D34%5E%7B1-2%7D%5C%5C%5C%5C%3D2%5Ccdot%2095%5E%7B-3%5Cfrac%7B1%7D%7B3%7D%7D%5Ccdot%2034%5E%7B-1%7D%3D2%5Ccdot%2095%5E%7B-4%2B%5Cfrac%7B2%7D%7B3%7D%7D%5Ccdot%2034%5E%7B-1%7D%5C%5C%5C%5C%3D%5Cdfrac%7B2%5Csqrt%5B3%5D%7B95%5E2%7D%7D%7B95%5E4%5Ccdot%2034%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B95%5E2%7D%7D%7B17%5Ccdot95%5E4%7D%5C%5C%5C%5C%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B9%5C%2C025%7D%7D%7B1%5C%2C384%5C%2C660%5C%2C625%7D)
