All Answers 358090 Answers

Gregory planted a lemon tree in hid back yard. When he planted the tree, it was 2 feet tall. He noticed that it has been growing 3 inches every week. Create multiple representations (x to y table, graph and equation) to represent the relationship between the days that have passed and the height of the tree.

I'd do this a bit different from Chris.  The question refers to growth in inches and time in weeks. So they are the units I would use.  Time along the bottom and height on the horizonal axis.

$$H=24+3T\\
\begin{tabular}{|c|c|c|c|c|c|} \hline
$Time\; in\; Week$&0&1&2&3&4\\\hline
$Height\; in\; Inches$ &24&27&&&\\\hline
\end{tabular}\\\\
\mbox{After one week the tree has grown 3 inches. }\\
\mbox{So it will be }24+3=27 \mbox{ inches high. etc}$$

"...get the inverse matrix and then multiply it with the constants matrix to get the solution to the systems of equations problem but using a matrix instead of doing it by hand...."

You could do this all in one step here, by changing B to vector and entering the appropriate values, then changing the + sign to a * sign.

No! That doesn't work!  For some reason the calculator here tries to calculate A^-B if you do that!!  You'll have to do it in two separate stages (i.e. find A^-1 first then reenter that as a new A and multiply by B).

You can use the constant acceleration formula: v = u + at

v is  final velocity (0 m/s)

u is initial velocity (20 m/s)

a is acceleration (-4/3 m/s²: the minus sign shows it is decelerating)

t is the time (in seconds)

So:

0 = 20 - (4/3)*t

Rearrange:

(4/3)t = 20

t = 20*3/4

t = 15 seconds

Is  this you intended question?  

If it is, this is how you should have written it   (-4*5)/(1/5+(-0.8))  Please look and learn how to present your problems it is important and your questions are more likely to  be answered quickly if we all understand what you question actually is.

$$\frac{-4*5}{\frac{1}{5}+(-0.8)}\\\\
=\frac{-4*5}{0.2-0.8}\\\\
=\frac{-20}{-0.6}\\\\
=\frac{20}{0.6}\\\\
=\frac{200}{6}\\\\
=\frac{100}{3}\\\\
=33.\overline{3}\quad OR \quad 33\frac{1}{3}$$

I think our answers are the same aren't they?  I just like being long winded.  

I solve these and write the code at the same time.  That's probably why I get in such a mess and it takes so long.  But it is more fun that way.  

If those NASCAR tire changing people are there, it will take less than a minute!

That's one thing about these identities......there are many approaches, sometimes......

Sorry Chris, I can't resist.    

Your will be amused to know that I got in a mess and it took forever.   lol

$${\frac{{\mathtt{1}}}{\left({\mathtt{cosecA}}{\mathtt{\,-\,}}{\mathtt{cotA}}\right)}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{sinA}}}} = {\frac{{\mathtt{1}}}{{\mathtt{sinA}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{\left({\mathtt{cosecA}}{\mathtt{\,\small\textbf+\,}}{\mathtt{cotA}}\right)}}$$

$$\begin{array}{rll}
LHS&=&\frac{sinA}{sinA(cosecA-cotA)}-\frac{(cosecA-cotA)}{sinA(cosecA-cotA)}\\\\
&=&\frac{sinA-(cosecA-cotA)}{sinA(cosecA-cotA)}\\\\
&=&\frac{sinA-cosecA+cotA}{sinA(\frac{1}{sinA}-\frac{cosA}{sinA})}\\\\
&=&\frac{sinA-cosecA+cotA}{1-cosA}\\\\
&=&\frac{(sinA-cosecA+cotA)sinA}{(1-cosA)sinA}\\\\
&=&\frac{sin^2A-1+cosA}{(1-cosA)sinA}\\\\
&=&\frac{1-cos^2A-1+cosA}{(1-cosA)sinA}\\\\
&=&\frac{cosA(1-cosA)}{(1-cosA)sinA}\\\\
&=&\frac{cosA}{sinA}\\\\
&=& CotA\\\\\\
RHS&=&\frac{(cosecA+cotA)}{sinA(cosecA+cotA)}-\frac{sinA}{sinA(cosecA+cotA)}\\\\
&=&\frac{cosecA+cotA-sinA}{1+cosA}\\\\
&=&\frac{(1+cosA-sin^2A)}{(1+cosA)sinA}\\\\
&=&\frac{(1+cosA-(1-cos^2A)}{(1+cosA)sinA}\\\\
&=&\frac{cosA(1+cosA)}{(1+cosA)sinA}\\\\
&=&\frac{cosA}{sinA}\\\\
&=& cotA\\\\
&=& LHS \qquad QED\\\\







\end{array}$$

90% x 630,000 = 567,000 voted yes

The linear equation that models this is:

y = (3/7)x + 24    where y is the height of the tree on any given day in inches and x represents the number of days elapsed after day "0."  Note that, on day "0," the tree is 24 inches tall = 2ft.

Here's the graph here: 

P.S. - You can create a table on your own.....just plug a few values of x into the above equation and generate some different y values

1/[csc A - cotA] - 1/sinA = 1/sinA  - 1/[csc A + cot A]     get everything in terms of sine and cosine

1/ [(1/sinA - cosA/sinA)] - 1/sinA = 1/sinA - 1/ [(1/sinA + cosA/sinA)]    simplify this

sinA/[1 - cosA] - 1/sinA = 1/sinA - sinA/[1 + cosA]

Multiply the first term on the LHS by (1 + cos A) in the the numerator and the denominator. The denominator becomes (1-cos^A) = sin^2A.......the same sort of thing is also done to the last term on the RHS....

sinA[1 + cosA]/sin^2A - 1/sinA = 1/sinA - sinA[1-cosA]/sin^2A

Get common denominators - (sin^2A) - on both sides

sinA[1+cosA]/sin^2A -sinA/sin^2A  = sinA/sin^2A - sinA[1 - cosA]/sin^2a

sinA/sin^2a + cosA/sin^A - sinA/sin^2A = sinA/sin^2A - sinA/sin^2A + cosA/sin^2A

cosA/sin^2A    = cosA/sin^2A

Sounds like you have a computer virus, rosala

Go here and download malwarebytes....don't worry...it's safe!!

http://www.malwarebytes.org/

Once you've downloaded it.....run a scan...it will usually catch Trojan Horses, etc.

I use Adblock Edge to block ads...but I think you might have something else going on....

Thank you Aziz!ive just installed it!And thanks xvxv too!

Points for both of you!

have a nice day!

sorry for asking !  but I don't understand the first question 

You can use Adblock .. do you know how to install it ! 

for 2 .. I can't understand why v2 smaller than v1 

Rosala, get the extension for Google Chrome called Adblock, and it'll block all ads!

yes i mean those advertisments!

And what do u mean by what is your browser?

i am using Google Chrome as my default browser!

You mean advertisments when you browsing the net ! 

if yes what is your browser ! ? 

Parallel lines have the same slopes.

Just put 200^99 into the calc on the home page and I think you will get the answer that you want.

Im on a smart phone and it won't let me post it from here. 

$${{\mathtt{200}}}^{{\mathtt{99}}} = {\mathtt{633\,825\,300\,114\,114\,700\,748\,351\,602\,688\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000\,000}}$$

$${\mathtt{5.145\: \!6}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{{\mathtt{20}}} = {\mathtt{2\,496\,466\,029.500\: \!616\: \!854\: \!317\: \!836\: \!1}}$$

Assuming a mean of 0 and a standard deviation of 1, you would find the probablility values for corresponding z-scores in the back of your book in Table 1, A.1, etc. Anyway, it would be 0.3401.

There is no question.

You need to put your question and instructions together or you'll rarely get what you want.   We are not mind readers.

Beginning of 1st year 1

2nd year 200 

Beginning of 100th year 200^99

End of 100th year 200^100