So, since she weights 6 times more on earth, say for every lb on the moon is 6lbs on earth then.
now, if 1lb on the moon is 6lbs on earth, how much is 90 earth lbs on the moon?
Answer:
positive
Step-by-step explanation:
when there is no sign in front of the numbers, we can assume that the numbers are positive
Answer:
This is the equation for cellular respiration.
Step-by-step explanation:
C6H1206 is glucose. O2 is oxygen. They react to produce 6H20, which is water, and 6CO2, which is carbon dioxide. Energy is also produced which is represented by 2800 kJ. Energy is created in the form of ATP and heat.
Step-by-step explanation:
We can prove this by using similar triangles. We know that angle ABE and ACD are SIMILAR(we know that by the sign in between them).Therefore, we also know that CDA and BEA are similar, and EAB and DAC are similar. Since all the angles are similar, we also know that the side lengths are equal in proportion. AB/AE is equal to AC/AD. When assigned values, that would create a proportion that is equal since the smaller triangle is similar to the bigger one. If you want to create a proof, break each statement into its own section.
![\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\\\\ -----------------------------\\\\ 2cos(A)=3tan(A)\implies 2cos(A)=3\cfrac{sin(A)}{cos(A)} \\\\\\ 2cos^2(A)=3sin(A)\implies 2[1-sin^2(A)]=3sin(A) \\\\\\ 2-2sin^2(A)=3sin(A)\implies 2sin^2(A)+3sin(A)-2](https://tex.z-dn.net/?f=%5Cbf%20sin%5E2%28%5Ctheta%29%2Bcos%5E2%28%5Ctheta%29%3D1%5Cimplies%20cos%5E2%28%5Ctheta%29%3D1-sin%5E2%28%5Ctheta%29%0A%5C%5C%5C%5C%5C%5C%0Atan%28%5Ctheta%29%3D%5Ccfrac%7Bsin%28%5Ctheta%29%7D%7Bcos%28%5Ctheta%29%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%0A2cos%28A%29%3D3tan%28A%29%5Cimplies%202cos%28A%29%3D3%5Ccfrac%7Bsin%28A%29%7D%7Bcos%28A%29%7D%0A%5C%5C%5C%5C%5C%5C%0A2cos%5E2%28A%29%3D3sin%28A%29%5Cimplies%202%5B1-sin%5E2%28A%29%5D%3D3sin%28A%29%0A%5C%5C%5C%5C%5C%5C%0A2-2sin%5E2%28A%29%3D3sin%28A%29%5Cimplies%202sin%5E2%28A%29%2B3sin%28A%29-2)
![\bf \\\\\\ 0=[2sin(A)-1][sin(A)+2]\implies \begin{cases} 0=2sin(A)-1\\ 1=2sin(A)\\ \frac{1}{2}=sin(A)\\\\ sin^{-1}\left( \frac{1}{2} \right)=\measuredangle A\\\\ \frac{\pi }{6},\frac{5\pi }{6}\\ ----------\\ 0=sin(A)+2\\ -2=sin(A) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5C%5C%5C%5C%5C%5C%0A0%3D%5B2sin%28A%29-1%5D%5Bsin%28A%29%2B2%5D%5Cimplies%20%0A%5Cbegin%7Bcases%7D%0A0%3D2sin%28A%29-1%5C%5C%0A1%3D2sin%28A%29%5C%5C%0A%5Cfrac%7B1%7D%7B2%7D%3Dsin%28A%29%5C%5C%5C%5C%0Asin%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%3D%5Cmeasuredangle%20A%5C%5C%5C%5C%0A%5Cfrac%7B%5Cpi%20%7D%7B6%7D%2C%5Cfrac%7B5%5Cpi%20%7D%7B6%7D%5C%5C%0A----------%5C%5C%0A0%3Dsin%28A%29%2B2%5C%5C%0A-2%3Dsin%28A%29%0A%5Cend%7Bcases%7D)
now, as far as the second case....well, sine of anything is within the range of -1 or 1, so -1 < sin(A) < 1
now, we have -2 = sin(A), which simply is out of range for a valid sine, so there's no angle with such sine
so, only the first case are the valid angles for A
