![\bf \textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvertical%20parabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%204p%28y-%20k%29%3D%28x-%20h%29%5E2%20%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5Cqquad%20%5Cstackrel%7Bfocus~point%7D%7B%28h%2Ck%2Bp%29%7D%5Cqquad%20%5Cstackrel%7Bdirectrix%7D%7By%3Dk-p%7D%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22p%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22p%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
something noteworthy is that the squared variable is the "x", thus the parabola is a vertical one, the "p" value is negative, so is opening downwards, and the h,k is pretty much the origin,
vertex is at (0,0)
the focus point is "p" or 5 units down from there, namely at (0, -5)
the directrix is "p" units on the opposite direction, up, namely at y = 5
the focal width, well, |4p| is pretty much the focal width, in this case, is simply yeap, you guessed it, 20.
Answer:
Step-by-step explanation:
1) since the sixth term is 3 and the fifth term 24, the common ratio would be 3/24 = 1/8
The formula for finding the nth term of a geometric sequence is
Tn = ar^(n - 1)
If t6 = 3,r = 1/8, then
3 = a × 1/8^(6 - 1) = a × (1/8)^5
a = 3/(0.125)^5 = 98304
The first term is 98304.
Second term is 98304 × 1/8 = 12288
Third term is 12288 × 1/8 = 1536
Third term is 1536 × 1/8 = 192
2) t1 = 4
t2 = - 3t(2- 1) = - 3t1 = - 3 × 4 = - 12
t3 = - 3t(3- 1) = - 3t2 = - 3 × - 12 = 36
t4 = - 3t(4- 1) = - 3t3 = - 3 × 36 = - 108
3) let the numbers be t2,t3 and t4
The sequence becomes
1/2, t2,t3, t4,8
The formula for finding the nth term of a geometric sequence is
Tn = ar^(n - 1)
8 = 1/2 × r^(5 - 1)
8 = 1/2 × r^4
16 = r^4
2^4 = r^4
r = 2
t2 = 1/2 × 2 = 1
t3 = 1 × 2 = 2
t4 = 2 × 2 = 4
The answer in standard form is (x+5/2)^2+(y-2)^2=1
I completely forgot how to do this but maybe multiply the angles I’m so sorry
<h3><u>(2x - 5)(4x - 3)</u></h3>
The AC method, also known as splitting the middle, can be shown like this:
8x^2 - 26x + 15
<em><u>Check factors of 120.</u></em>
1 * 120
-1 * -120
2 * 60
-2 * -60
3 * 40
-3 * -40
5 * 24
-5 * -24
6 * 20
-6 * -20 (these factors, when added together, are equal to the middle term, and thus splitting the middle term is possible.)
<em><u>Split the middle term.</u></em>
8x^2 - 6x - 20x + 15
<em><u>Group in terms of 2.</u></em>
(8x^2 - 6x) - (20x + 15)
<em><u>Factor each binomial.</u></em>
2x(4x - 3) - 5(4x - 3)
<em><u>Rearrange the terms.</u></em>
(2x - 5)(4x - 3)
