Answer:
tanG= 15/20 = 3/4
tanX= 8/15
Step-by-step explanation:
tan is always opposite over adjacent
Slope is 2/-1 or -2. Because you go up twice and then you go left once.
Answer:
true
Step-by-step explanation:
When Ivan divides his coins into groups of 2 nickels and 1 quarter, worth $0.35, he find that he has $8.75/$0.35 = 25 such groups.
Ivan has 25 quarters and 50 nickels.
By the definition of the hyperbolic function tanh x, we have proven that ![\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}](https://tex.z-dn.net/?f=%5Cfrac%7B1-tanh%20%5C%20x%7D%7B1%20%2B%20tanh%20%5C%20x%7D%3De%5E%7B-2x%7D)
<h3>Hyperbolic functions & Proof of identities </h3>
By definition
![tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}](https://tex.z-dn.net/?f=tanh%20%5C%20x%3D%5Cfrac%7Be%5E%7Bx%7D%20-e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D)
Then,
![\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }](https://tex.z-dn.net/?f=%5Cfrac%7B1-tanh%20%5C%20x%7D%7B1%20%2B%20tanh%20%5C%20x%7D%3D%5Cfrac%7B1-%5Cfrac%7Be%5E%7Bx%7D%20-e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%7D%7B1%2B%20%5Cfrac%7Be%5E%7Bx%7D%20-e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%7D)
![=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })](https://tex.z-dn.net/?f=%3D1-%5Cfrac%7Be%5E%7Bx%7D%20-e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%5Cdiv%20%281%2B%20%5Cfrac%7Be%5E%7Bx%7D%20-e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%7D%29)
![=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })](https://tex.z-dn.net/?f=%3D%5Cfrac%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D-%28e%5E%7Bx%7D%20-e%5E%7B-x%7D%29%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%5Cdiv%20%28%5Cfrac%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%2Be%5E%7Bx%7D%20-e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%7D%29)
![=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D-e%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%5Cdiv%20%5Cfrac%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%2Be%5E%7Bx%7D%20-e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%7D)
![=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7Be%5E%7B-x%7D%2Be%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%5Cdiv%20%5Cfrac%7Be%5E%7Bx%7D%20%2Be%5E%7Bx%7D%20%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%7D)
![=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%5Cdiv%20%5Cfrac%7B2e%5E%7Bx%7D%20%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%7D)
![=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2e%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%20%5Ctimes%20%5Cfrac%7Be%5E%7Bx%7D%20%2Be%5E%7B-x%7D%7D%7B2e%5E%7Bx%7D%7D)
![=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2e%5E%7B-x%7D%7D%7B1%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2e%5E%7Bx%7D%7D)
![=\frac{2e^{-x}}{2e^{x}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2e%5E%7B-x%7D%7D%7B2e%5E%7Bx%7D%7D)
![=\frac{e^{-x}}{e^{x}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Be%5E%7B-x%7D%7D%7Be%5E%7Bx%7D%7D)
![=e^{-x} \times \frac{1}{e^{x}}](https://tex.z-dn.net/?f=%3De%5E%7B-x%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7Be%5E%7Bx%7D%7D)
![= e^{-x} \times e^{-x}](https://tex.z-dn.net/?f=%3D%20e%5E%7B-x%7D%20%5Ctimes%20e%5E%7B-x%7D)
![= e^{-x+-x}](https://tex.z-dn.net/?f=%3D%20e%5E%7B-x%2B-x%7D)
![= e^{-x-x}](https://tex.z-dn.net/?f=%3D%20e%5E%7B-x-x%7D)
![= e^{-2x}](https://tex.z-dn.net/?f=%3D%20e%5E%7B-2x%7D)
Hence, we have proven that ![\frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}](https://tex.z-dn.net/?f=%5Cfrac%7B1-tanh%20%5C%20x%7D%7B1%20%2B%20tanh%20%5C%20x%7D%3De%5E%7B-2x%7D)
Learn more on Proof of Identities here: brainly.com/question/2561079
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