I’m sorry, but I’m also lost looking at that .
oh my
Circle = (x-h)^2 + (y-k)^2 = r^2
Center is (h,k) h = -5, k = 2
Radius is 4, r = 4
(x - -5)^2 + (y - 2)^2 = 4^2
(x + 5)^2 + (y - 2)^2 = 16
Squaring a number is the same as multiplying the number by itself <span>(<span><span>64⋅64</span>).</span></span><span> In this case, </span>64<span> squared is </span><span>4096.
</span><span>8<span>r^3</span>−4096+r−8
</span>Subtract 8<span> from </span><span>−4096</span><span> to get </span><span><span>−4104</span>.
</span><span>8<span>r^3</span>−4104+r
</span>Factor<span> the </span>polynomial<span> using the rational </span>roots<span> theorem.
</span><span>(r−8)(8<span>r^2</span>+64r+513)</span>
False; consider as a counterexample the function <em>f</em> : ℝ→ℝ defined by
![f(x)=\begin{cases}-3&\text{for }x\neq2\\0&\text{for }x=2\end{cases}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cbegin%7Bcases%7D-3%26%5Ctext%7Bfor%20%7Dx%5Cneq2%5C%5C0%26%5Ctext%7Bfor%20%7Dx%3D2%5Cend%7Bcases%7D)
Clearly <em>f</em> approaches -3 as <em>x</em> gets closer to -2, but neither limit from either side is equal to the function's value at <em>x</em> = 2 (that is, -3 ≠ 0), so <em>f</em> is not continuous.
There are infinitely many ways to do this. One such way is to draw a very thin stretched out rectangle (say one that is very tall) and a square. Example: the rectangle is 100 by 2, while the square is 4 by 4.
Both the rectangle and the square have the same corresponding angle measures. All angles are 90 degrees.
However, the figures are not similar. You cannot scale the rectangle to have it line up with the square. The proportions of the sides do not lead to the same ratio
100/4 = 25
2/4 = 0.5
so 100/4 = 2/4 is not a true equation. This numerically proves the figures are not similar.
side note: if you are working with triangles, then all you need are two pairs of congruent corresponding angles. If you have more than three sides for the polygon, then you'll need to confirm the sides are in proportion along with the angles being congruent as well.