This is indeed a linear equation. y-4x=9 in standard form is y=4x+9.
Check the picture below, so it reaches the maximum height at the vertex, let's check where that is
![h(t)=64t-16t^2+0 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 64}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (64)^2}{4(-16)}\right)\implies \stackrel{maximum~height}{(2~~,~~\stackrel{\downarrow }{64})}](https://tex.z-dn.net/?f=h%28t%29%3D64t-16t%5E2%2B0%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ctextit%7Bvertex%20of%20a%20vertical%20parabola%2C%20using%20coefficients%7D%20%5C%5C%5C%5C%20h%28t%29%3D%5Cstackrel%7B%5Cstackrel%7Ba%7D%7B%5Cdownarrow%20%7D%7D%7B-16%7Dt%5E2%5Cstackrel%7B%5Cstackrel%7Bb%7D%7B%5Cdownarrow%20%7D%7D%7B%2B64%7Dt%5Cstackrel%7B%5Cstackrel%7Bc%7D%7B%5Cdownarrow%20%7D%7D%7B%2B0%7D%20%5Cqquad%20%5Cqquad%20%5Cleft%28-%5Ccfrac%7B%20b%7D%7B2%20a%7D~~~~%20%2C~~~~%20c-%5Ccfrac%7B%20b%5E2%7D%7B4%20a%7D%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28-%5Ccfrac%7B%2064%7D%7B2%28-16%29%7D~~~~%20%2C~~~~%200-%5Ccfrac%7B%20%2864%29%5E2%7D%7B4%28-16%29%7D%5Cright%29%5Cimplies%20%5Cstackrel%7Bmaximum~height%7D%7B%282~~%2C~~%5Cstackrel%7B%5Cdownarrow%20%7D%7B64%7D%29%7D)
All correct accept "man" the word "man" goes to the long a side
you are all good
hopes this help :) :D :)
Answer:
The correct answer is - 7.166 years
Step-by-step explanation:
Given:
principle amount: 3000
rate of interest: 2%
time?
Interent to get: 430
Formula:
I = P*t*r/100
here p = principle
I = interest
r = rate of intrest and t = time
Solution putting value and deriving Time as formula:
(3000*2*t)/100 = 430
t = 43000/3000*2
= 7.166 years.
The answer would be 17/20
