Y = 2x + 2
Y = 2x -
Subtract
0 = 2
This makes no sense meaning the lines are parallel. Therefore there are 0 solutions.
Check the picture below.
now, let's notice the larger "yellow" semicircle, it has a gap, the gap on the right is of a semicircle with a diameter of 10, BUT it also has a descender on the left, a part that's hanging out, that part is also a semicircle.
so if we use the descending semicircle to fill up the gap on the right, we'll end up with a filled up larger semicircle, whose diameter is 20, and whose radius is 10 cm.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=10 \end{cases}\implies A=\pi 10^2\implies A=100\pi \\\\\\ \stackrel{\textit{half of that for a semicircle}}{A=\cfrac{100\pi }{2}}\implies A=50\pi \implies \stackrel{\pi =3.142}{A=157.1}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D10%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%2010%5E2%5Cimplies%20A%3D100%5Cpi%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bhalf%20of%20that%20for%20a%20semicircle%7D%7D%7BA%3D%5Ccfrac%7B100%5Cpi%20%7D%7B2%7D%7D%5Cimplies%20A%3D50%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Cpi%20%3D3.142%7D%7BA%3D157.1%7D)
There aren't any specified coordinates?
I'll use multiples of 2 and 4 as an example:
Multiples of 2: 2, 4, 6, 8...
Multiples of 4: 4, 8, 12, 16...
The least common multiple in this case is 4. The LCM is always ≥ the largest starting number, which is 4 for this example. Therefore, the statement is true.
<em>Hope this helps! :)</em>
Answer:
b = -T²/k(k-m) + k
Step-by-step explanation:
Given the equation T =√k(k-b)(k-m), we are to make b the subject of the formula
Square both sides
T² = (√k(k-b)(k-m))²
T² = k(k-b)(k-m)
T²/k(k-m) = k-b
Rearrange
k - b = T²/k(k-m)
b - k = - T²/k(k-m)
Add j=k to both sides
b - k + k = - T²/k(k-m) + k
b = -T²/k(k-m) + k
This gives the required expression
