<h2>Answer: y = x - 1</h2>
<h3>Step-by-step explanation:
</h3>
<u>Find the slope of the parallel line</u>
When two lines are parallel, they have the same slope.
⇒ if the slope of this line = 1
then the slope of the parallel line (m) = 1
<u>Determine the equation</u>
We can now use the point-slope form (y - y₁) = m(x - x₁)) to write the equation for this line: (5, 4)
⇒ y - 4 = 1 (x - 5)
∴ y - 4 = x - 5
We can also write the equation in the slope-intercept form by making y the subject of the equation and expanding the bracket to simplify:
since y - 4 = x - 5
<h3> y = x - 1</h3>
To test my answer, I drew a Desmos Graph with the information provided in the question and my answer.
Answer:
Yes, you can use similar triangles to find x
Answer:
![(\sqrt{2})^3](https://tex.z-dn.net/?f=%28%5Csqrt%7B2%7D%29%5E3)
Step-by-step explanation:
Given the expression
![(x - \frac{1}{x} )^3](https://tex.z-dn.net/?f=%28x%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%29%5E3)
Simplify
![(\frac{x^2-x}{x} )^3](https://tex.z-dn.net/?f=%28%5Cfrac%7Bx%5E2-x%7D%7Bx%7D%20%29%5E3)
Given that x = 1+√2
Substitute
![(\frac{(1+\sqrt{2} )^2-(1+\sqrt{2} )}{1+\sqrt{2} } )^3\\=( \frac{1+2\sqrt{2} +2 - 1 - \sqrt{2}}{1+\sqrt{2}} )^3\\= (\frac{\sqrt{2}+2}{1+\sqrt{2}})^3 \\Rationalize\\= (\frac{\sqrt{2}-2+2-2\sqrt{2}}{1-2})^3 \\=( \frac{-\sqrt{2}}{-1})^3 \\= (\sqrt{2})^3\\ \\](https://tex.z-dn.net/?f=%28%5Cfrac%7B%281%2B%5Csqrt%7B2%7D%20%29%5E2-%281%2B%5Csqrt%7B2%7D%20%29%7D%7B1%2B%5Csqrt%7B2%7D%20%7D%20%29%5E3%5C%5C%3D%28%20%5Cfrac%7B1%2B2%5Csqrt%7B2%7D%20%2B2%20-%201%20-%20%5Csqrt%7B2%7D%7D%7B1%2B%5Csqrt%7B2%7D%7D%20%29%5E3%5C%5C%3D%20%28%5Cfrac%7B%5Csqrt%7B2%7D%2B2%7D%7B1%2B%5Csqrt%7B2%7D%7D%29%5E3%20%5C%5CRationalize%5C%5C%3D%20%28%5Cfrac%7B%5Csqrt%7B2%7D-2%2B2-2%5Csqrt%7B2%7D%7D%7B1-2%7D%29%5E3%20%5C%5C%3D%28%20%5Cfrac%7B-%5Csqrt%7B2%7D%7D%7B-1%7D%29%5E3%20%5C%5C%3D%20%28%5Csqrt%7B2%7D%29%5E3%5C%5C%20%5C%5C)