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What you need to do is realise that the line that goes through (1,6) and (4,4) has a gradient of -2/3 (using rise/run). Any line perpendicular to this will have a negative reciprocal gradient, as perpendicular lines' gradient multiply together to give -1. The gradient of the line will thus be 3/2. All you have to do is find the equation of the line that has a gradient of 3/2 and also passes through (-2,6), and sketch it.

OK.....   can you tell me what the equation for the line is?

See if you can figure THAT out and repost on this thread for more help..... 

Here is ANOTHER little trick...THIS website IS a calculator.....go to the homesite and type in 2.333333 and

hit ' = '

and look above the calc display to see 7/3         F'in'     majic.     Try it !

Girls   12/30       Boys   16/40

Does 12/30 = 16/40  ?

          2/5     = 2/5      Yes !     The ratios are the same !   (Tertre doesn't usually miss these Q's !)

Yup!    I would vote for Lucas' parents for being the better....but only barely.   Grrrrrr........

\(t_1 = 1\\ t_{n+1} = t_n + n \\ t_2 = 1 + 2 = 3\\ t_3 = 3 + 3 = 6 \\ t_4 = 6 + 4 = 10 \\ t_5 = 10 + 5 = 15\)

This matches up with choice (c)

I did copy the question correct, thank you for the help :) 

sqrt(48x) = 24 sqrt(3)       Square both sides

48x = 576 (3)

x=36

PLEASE post your questions HERE....not LINKS.....

How do I do this?

Evaluate: f(2) and f(-1) and f(100) for f(x)=-5+35

I expect you have not copied your question properly.

But on the off chance that you didn't, f(x)=30 no matter what the value of x is.

so f(2)=30 and f(-1)=30

This question is actually much simpler than the others that you have presented.

Just use the basic angle rules for parallel lines and then concentrate on the little triangle at the top.

Maybe let angle 

Hi Shreyas

Tell you what.

I (or someone else) will probably have time to do this  a little later if you present it properly.

You can edit your original question since you are not actually changing it. You are just makeing it readable.

I'll show you how.

Here is a bit of your question.

$\overline{ST}\parallel\overline{QR}$

Depending on how your computer is set up this may present properly to you but it does not present properly to me or to most people.

1) you need to get rid of the dollar signs

\overline{ST}\parallel\overline{QR}

2) Then you need to copy it

3) then you need to open the Latex button that is in the ribbon.

4) then you need to paste it into the LaTex box and press return.

5) it should now present properly underneath.

              \(\overline{ST}\parallel\overline{QR}\)      

6) click out of the latex box and it should present properly now.

             In the diagram below, we have  \(\overline{ST}\parallel\overline{QR}\)      etc

7) Do the same thing for all the other bits of latex code that is in the question.

If you cannot make this work then pester us until you have enough info to work it out.

It is important that you learn to do this and it is not very difficult.

Please no one answer this question until Shreyas repairs it.  She/he needs to learn how to present her/his questions properly.

here's a little trick

\(9x = 10x - 1x \\ 9(2.333\bar{3}) = 10(2.333\bar{3}) - 2.333\bar{3} = \\ 23.333\bar{3} - 2.333\bar{3} = 21 \\ x = \dfrac{21}{9} = \dfrac 7 3\)

Bradley has purchased a home for $471,000. He made a 25% down payment and financed the remaining amount. The intangible tax is 0.5%. Which of the following is the total amount of the intangible tax? (2 points)

$176,625.00

$1,766.25

$235,500.00

Hint: Use the lines, that lie on each side of the triangle! They mean that the angles are congruent!

We have two ratios. 

Girls: \(12:40=3:10\)

And, for the boys: \(16:40=4:10=2:5\).

Thus, the answer is \(\boxed{NO}\).

never mind I got the answer

\(\text{Assume they both spent }N \text{ hours doing homework}\\ V_J = \dfrac{N}{5}\times 3 = \dfrac 3 5 N\\ V_L = \dfrac 1 2 N \\ \dfrac 3 5 > \dfrac 1 2 \\ \text{Jonathan gets to spend more time playing video games}\)

\(\text{Consider a win on the nth roll.}\\ \text{This roll sequence looks like }\\ \underset{n_T}{\underbrace{TTT\dots T}}\underset{n_H}{\underbrace{HHH\dots H}}T\\ 0\leq n_T \leq n-2,~1\leq n_H \leq n-1,~n_T+n_H +1 = n\)

\(\text{Thus there are }n-1 \text{ ways to accomplish a win on the nth roll}\\ \text{Each sequence has probability }p_n = 2^n \text{thus the probability of a win on the nth roll is given by}\\ P[n] = \dfrac{n-1}{2^n},~~n\in \mathbb{N},~2 \leq n\)

\(E[N] = \sum \limits_{n=2}^\infty~\dfrac{n(n-1)}{2^n}=6-2= 4\)

You can apply the same idea to solve 2 and 3

Multiply the second term by   sqrt(n - 1)  /  Sqrt(n-1)

then you would have    { 1  +  (n-1)  }  / sqrt(n-1) ===   n / sqrt(n-1)    

Thanx , Chris !   Guess I should not have changed mt original answer......

Here's my attempt at this.....

For the in-divisional games  we can choose any 2 of 7 teams in any division  = (7, 2)  = 21 unique  pairings

Since each unique pair plays twice....this gives 2 * 21 = 42 games played within one division

And since there are 2  divisions...this gives us  2 * 42   = 84  in-divsional games

And since every team in plays every team in the other divsion once this gives us 7 * 7  = 49 more games

So...the total games played are  84 + 49    =  133 games

Your answer of "D" is correct...!!!

ik but I put it on the off topic did I not

Not sure anyone in their right mind would answer this on a MATH forum......C'mon.....This is not a dating or stalking site....!

g(3) = -2(3)+1 = -5    now substitute this in to the  f(x) function

f(-5) = -5^3 + 4(-5) = -125 - 20 = -145      "B"