The smaller wheel rolled
.. 4*2πr = 8πr . . . . feet
The larger wheel rolled
.. 4*2π*2r = 16πr . . feet
The larger wheel traveled (16πr -8πr) = 8πr feet farther than the small wheel.
Basically, you just move every individual point up, down, left, or right by the amount indicated.
For example, point G is graphed on the point (-3, -1). Moving it right 5 and up 1 will give you G’, which is (2,0)
T is graphed on the point (-1, -1). Moving it right 5 and up 1 will give you T’, which is (4,0)
B is graphed on the point (-3, -5). Moving it right 5 and up 1 will give you B’, which is
(2,4)
Check the picture below, so the hyperbola looks more or less like so, so let's find the length of the conjugate axis, or namely let's find the "b" component.
![\textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Ctextit%7Bhyperbolas%2C%20horizontal%20traverse%20axis%20%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B%28x-%20h%29%5E2%7D%7B%20a%5E2%7D-%5Ccfrac%7B%28y-%20k%29%5E2%7D%7B%20b%5E2%7D%3D1%20%5Cqquad%20%5Cbegin%7Bcases%7D%20center%5C%20%28%20h%2C%20k%29%5C%5C%20vertices%5C%20%28%20h%5Cpm%20a%2C%20k%29%5C%5C%20c%3D%5Ctextit%7Bdistance%20from%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bcenter%20to%20foci%7D%5C%5C%20%5Cqquad%20%5Csqrt%7B%20a%20%5E2%20%2B%20b%20%5E2%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
That's false.
For similar triangles, corresponding angles are congruent.
Corresponding sides are in the same ratio but rarely congruent.
(They're congruent only if the ratio is ' 1 ', i.e. the triangles are congruent
as well as being similar.)
Answer:
Answer and explanation with steps are in the following attachments
Step-by-step explanation:
