E. he did not multiply 3 by -1 correctly
The simplify expression is 9 - 3 + 6x = 6 + 6x
A = base side=3
b= base side=5
c = base side=3
h= height=7-4=3
l= lenght=6
w= width=5
h= height=4
V= 1/4h√(-a⁴+2(ab)²+2(ac)²-b⁴+2(bc)²-c⁴) + l w h
V=12.44+ 120=132.44 unit³
Answer:
![\displaystyle 1)48.2 \: \: \text{sec}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%201%2948.2%20%20%20%20%5C%3A%20%20%5C%3A%20%5Ctext%7Bsec%7D)
![\rm \displaystyle 2)3021.6 \: m](https://tex.z-dn.net/?f=%20%5Crm%20%5Cdisplaystyle%20%202%293021.6%20%5C%3A%20m)
Step-by-step explanation:
<h3>Question-1:</h3>
so when <u>flash down</u><u> </u>occurs the rocket will be in the ground in other words the elevation(height) from ground level will be 0 therefore,
to figure out the time of flash down we can set h(t) to 0 by doing so we obtain:
![\displaystyle - 4.9 {t}^{2} + 229t + 346 = 0](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20%20-%204.9%20%7Bt%7D%5E%7B2%7D%20%20%2B%20229t%20%2B%20346%20%3D%200)
to solve the equation can consider the quadratic formula given by
![\displaystyle x = \frac{ - b \pm \sqrt{ {b}^{2} - 4 ac} }{2a}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20x%20%3D%20%20%5Cfrac%7B%20-%20b%20%5Cpm%20%20%5Csqrt%7B%20%7Bb%7D%5E%7B2%7D%20-%204%20ac%7D%20%7D%7B2a%7D%20)
so let our a,b and c be -4.9,229 and 346 Thus substitute:
![\rm\displaystyle t = \frac{ - (229) \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}](https://tex.z-dn.net/?f=%20%20%5Crm%5Cdisplaystyle%20t%20%3D%20%20%5Cfrac%7B%20-%20%28229%29%20%5Cpm%20%20%5Csqrt%7B%20%7B229%7D%5E%7B2%7D%20-%204.%28%20-%204.9%29%28346%29%7D%20%7D%7B2.%28%20-%204.9%29%7D%20)
remove parentheses:
![\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}](https://tex.z-dn.net/?f=%20%20%5Crm%5Cdisplaystyle%20t%20%3D%20%20%5Cfrac%7B%20-%20229%20%5Cpm%20%20%5Csqrt%7B%20%7B229%7D%5E%7B2%7D%20-%204.%28%20-%204.9%29%28346%29%7D%20%7D%7B2.%28%20-%204.9%29%7D%20)
simplify square:
![\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 4( - 4.9)(346)} }{2.( - 4.9)}](https://tex.z-dn.net/?f=%20%20%5Crm%5Cdisplaystyle%20t%20%3D%20%20%5Cfrac%7B%20-%20229%20%5Cpm%20%20%5Csqrt%7B%2052441-%204%28%20-%204.9%29%28346%29%7D%20%7D%7B2.%28%20-%204.9%29%7D%20)
simplify multiplication:
![\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 6781.6} }{ - 9.8}](https://tex.z-dn.net/?f=%20%20%5Crm%5Cdisplaystyle%20t%20%3D%20%20%5Cfrac%7B%20-%20229%20%5Cpm%20%20%5Csqrt%7B%2052441-%206781.6%7D%20%7D%7B%20-%209.8%7D%20)
simplify Substraction:
![\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 45659.4} }{ - 9.8}](https://tex.z-dn.net/?f=%20%20%5Crm%5Cdisplaystyle%20t%20%3D%20%20%5Cfrac%7B%20-%20229%20%5Cpm%20%20%5Csqrt%7B%2045659.4%7D%20%7D%7B%20-%209.8%7D%20)
by simplifying we acquire:
![\displaystyle t = 48.2 \: \: \: \text{and} \quad - 1.5](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20t%20%3D%2048.2%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%5Ctext%7Band%7D%20%5Cquad%20%20-%201.5)
since time can't be negative
![\displaystyle t = 48.2](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20t%20%3D%2048.2%20%20)
hence,
at <u>4</u><u>8</u><u>.</u><u>2</u><u> </u>seconds splashdown occurs
<h3>Question-2:</h3>
to figure out the maximum height we have to figure out the maximum Time first in that case the following formula can be considered
![\displaystyle x _{ \text{max}} = \frac{ - b}{2a}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20x%20_%7B%20%20%5Ctext%7Bmax%7D%7D%20%3D%20%20%5Cfrac%7B%20-%20b%7D%7B2a%7D%20)
let a and b be -4.9 and 229 respectively thus substitute:
![\displaystyle t _{ \text{max}} = \frac{ - 229}{2( - 4.9)}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20t%20_%7B%20%20%5Ctext%7Bmax%7D%7D%20%3D%20%20%5Cfrac%7B%20-%20229%7D%7B2%28%20-%204.9%29%7D%20)
simplify which yields:
![\displaystyle t _{ \text{max}} = 23.4](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20t%20_%7B%20%20%5Ctext%7Bmax%7D%7D%20%3D%20%2023.4)
now plug in the maximum t to the function:
![\rm \displaystyle h(23.4)- 4.9 {(23.4)}^{2} + 229(23.4)+ 346](https://tex.z-dn.net/?f=%20%5Crm%20%5Cdisplaystyle%20%20h%2823.4%29-%204.9%20%7B%2823.4%29%7D%5E%7B2%7D%20%20%2B%20229%2823.4%29%2B%20346%20)
simplify:
![\rm \displaystyle h(23.4) = 3021.6](https://tex.z-dn.net/?f=%20%5Crm%20%5Cdisplaystyle%20%20h%2823.4%29%20%20%3D%20%203021.6)
hence,
about <u>3</u><u>0</u><u>2</u><u>1</u><u>.</u><u>6</u><u> </u>meters high above sea-level the rocket gets at its peak?