Answer:
C. The cost of a box of cereal
Step-by-step explanation:
In the question, we are given the number of 3 boxes of cereal, 2 loaves of bread, and that the total cost is $15. We can see these numbers in the equation, and putting it into context, we see that the 15 is on the right side of the equation, which indicates cash. Therefore, the 3 and 2 are accounted for, but their prices are unknown. X is a placeholder for the unidentified cost of a box of cereal, and we know that it's the boxes of cereal and not the loaves of bread because there are 3 and not 2.
Answer: she biked 0.8 hour each day
Step-by-step explanation:
Let x represent the number of hours that she jogged on both days.
Let y represent the number of hours that she biked on both days.
Distance = speed × time
On the first day, she jogs at a rate of 6 miles/hour and bikes at a rate of 10 miles/hour. She covers a total of 26 miles that day. This is expressed as
6x + 10y = 26- - - - - - - - - - - - - - - 1
The second day, she continues to jog at a rate of 6 miles/hour, but a hamstring pull slows her biking rate to 5 miles/hour. The second day she covers a total of 22 miles. This means that
6x + 5y = 22- - - - - - - - - - - -2
Subtracting equation 2 from equation 1, it becomes
5y = 4
y = 4/5 = 0.8
Substituting y = 0.8 into equation 1, it becomes
6x + 10(0.8) = 26
6x + 8 = 26
6x = 26 - 8 = 18
x = 18/6
x = 3
Answer:
Step-by-step explanation:
Given
Required
Show that these two lines have no point of intersection
The general equation of a line is
Where m represents slope
In
Similarly; in
Since the slope of both lines are equal;
<em>This implies that the lines are parallel and parallel lines have no point of intersection</em>
<em></em>
<em>See Attachment for Graph</em>
Answer:
13/9 or 1 4/9
Step-by-step explanation:
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)
