Amount of liquid plant fertilizer in the container that the gardener has = 6 1/5 pounds
= 31/5 pounds
We already know that 1 pound = 16 ounce
Then
Amount of liquid plant fertilizer in the
container that the gardener has = (31/5) * 16 ounce
= 496/5 pounds
Amount of liquid plant fertilizer that the gardener uses on Sunday = 2 1/2 ounces
= 5/2 ounces
Then
The amount of liquid plant fertilizer that are left= (496/5) - (5/2) ounces
= (992 - 25)/10 ounces
= 967/10 ounces
= 96.7 ounces
So the total amount of liquid plant fertilizer left is 96.7 ounces.
Absolute value is always positive so I would say the answer would be A.
-38(-3) = 114
-72/(-12) = 6
-9 x 23 = -207
-150/5 = -30
564 / -4 = -141
<h3><u>Solution:</u></h3>
Given that, we have to find each product or quotient
<em><u>-38(-3)</u></em>
It is a product, so we have to find product value

<em><u>-72 / (-12)</u></em>
It is division, so we have to find the quotient

<em><u>-9 x 23</u></em>
It is a product, so we have to find product value

<em><u>- 150 / 5</u></em>
It is division, so we have to find quotient

<em><u>564 / -4</u></em>
It is division, so we have to find the quotient

Check the picture below.
where is the -16t² coming from? that's Earth's gravity pull in feet.
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{30}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{6}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+30t+6 \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B30%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B6%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20h%28t%29%3D-16t%5E2%2B30t%2B6%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

![\bf \left(-\cfrac{30}{2(-16)}~~,~~6-\cfrac{30^2}{4(-16)} \right)\implies \left( \cfrac{30}{32}~,~6+\cfrac{225}{16} \right)\implies \left(\cfrac{15}{16}~,~\cfrac{321}{16} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\stackrel{\textit{how many}}{\textit{seconds it took}}}{0.9375}~~,~~\stackrel{\stackrel{\textit{how many feet}}{\textit{up it went}}}{20.0625})~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%28-%5Ccfrac%7B30%7D%7B2%28-16%29%7D~~%2C~~6-%5Ccfrac%7B30%5E2%7D%7B4%28-16%29%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%20%5Ccfrac%7B30%7D%7B32%7D~%2C~6%2B%5Ccfrac%7B225%7D%7B16%7D%20%5Cright%29%5Cimplies%20%5Cleft%28%5Ccfrac%7B15%7D%7B16%7D~%2C~%5Ccfrac%7B321%7D%7B16%7D%20%5Cright%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%28%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bhow%20many%7D%7D%7B%5Ctextit%7Bseconds%20it%20took%7D%7D%7D%7B0.9375%7D~~%2C~~%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bhow%20many%20feet%7D%7D%7B%5Ctextit%7Bup%20it%20went%7D%7D%7D%7B20.0625%7D%29~%5Chfill)
Answer:
It is -5
Step-by-step explanation: