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\(\Delta DEB\)  is a \(30 - 60 - 90\) triangle, so \(BE = \sqrt{3}(DE)\). \(EC = DE\), and since we are only looking for ratios, we can assign \(DE\) and \(EC\) to \(x\). \(\Delta ABC\) is a \(30 - 60 - 90\) triangle also. \(AC\) is \(4x\) because \(BD\) is a median, therefore splitting \(AC\) into two. Since \(AC = 4x\), \(BC = 2x\). (If you didn't know already, a \(30 - 60 - 90\) triangle's sides are in a ratio of \(1 : \sqrt{3} : 2\).) If \(BC = 2x\), then \(AB = 2\sqrt{3}x\), and \(EC = x\), so your ratio would be \(\frac{2\sqrt{3}x}{x}\), which simplifies to \(2\sqrt{3}\) when you cancel the \(x\)'s.

- Daisy

OK....  since M is the mid-point of  QR then MR  =  17

This triangle is isosceles so  if  we let QS  be the altitude....then this altitude will bisect PR..

So....SR  = 16

And  the  cosine of angle QRS  = SR/ QR  = 16/34  =  8/17

Using the Law of  Cosines

PM^2  = MR^2  + PR^2  - 2(MR)(PR)cos(QRS)

PM^2  = 17^2 + 32^2  - 2(17)(32) (8 /17)

PM^2  = 1313  - 16*32

PM^2 = 1313 - 512

PM  = sqrt (1313 - 512)

PM =sqrt (801)  = sqrt ( 9 * 89)  =  3  sqrt (89) ≈ 28.3  units

Here's a pic :

Where is a dislike button when you need one?

Wow, that is amazing! Congrats you got both the loan balance and the annual cost correct! I don't think I have seen that scary formula, but I did try to calculate one of the loan payments and it worked. I will spend some time to see how you solved it in detail. I thank you for the time you took to do this.

Similarly...for the second one...let's pick  ( 4 , 0)  = (x , y)

Look  at the last answer..... plug in  4 for x   and 0  for y  and we have

0 > -2(4) - 2

0 > -8 - 2

0 >  -10     which is true

Also

0  ≤ -(1/2)(4) + 6

0 ≤ -2 + 6

0 ≤ 4     which is also true

Lastly

0 ≤ 4     which is also true

Since  all  three inequalities are true, this must be the correct answer

Check for yourself that at least one equation in each of  the other three answers gives an incorrect inequality

For example in the first two answers

y ≤ -2x + 6     will not be true for  (x, y)  =  (4,0)

And in the third answer

y ≥ x  will not be true  for  (x, y)  = ( 4,0)

Hope that helps.....!!!!

First one...here's one way to do this....pick a point in the shaded area...I'll choose (1,2)  = (x , y)

Note that this satisfies   y  ≥   1      and    y  - x  >  0

So...the second answer is correct for the first one

(a - b)^3 + (b - c)^3  + ( c -a)^3  =

[( a - b ) + (b - c)] [ ( a - b)^2 - ( a - b)(b - c) + ( b - c)^2 ] + (c -a)^3 =

( a - c) [ a^2 - 2ab + b^2  - [ ab - b^2 - ac + bc] + b^2 - 2bc + c^2 ] + (c -a)^3 =

(a - c)  [ a^2 - 2ab + b^2  - ab + b^2 + ac- bc  + b^2 - 2bc + c^2 ]  + (c -a)^3  =

(a - c) [ a^2 - 3ab + 3b^2 + ac - 3bc  + c^2 ]  + (c - a)^3 =

(a - c) [ a^2 - 3ab + 3b^2 + ac - 3bc  + c^2 ] + ( c - a)^3]=

(a - c) [ a^2 - 3ab + 3b^2 + ac - 3bc  + c^2 ] + [ (c - a) (c - a)^2 ]  =

- (c - a) [ a^2 - 3ab + 3b^2 + ac - 3bc + c^2 ] [ (c - a)( c^2 - 2ac + a^2 ]  =

(c - a) [ -a^2 + 3ab - 3b^2 - ac + 3bc - c^2 + c^2 - 2ac + a^2 ] =

(c -a) [ 3ab  - 3b^2 - 3ac + 3bc] =

(c - a) [ 3ab - 3ac - 3b^2 + 3bc ] =

(c - a) [ 3a ( b - c) - 3b ( b - c) ]  =

(c - a) [3(a - b)) ( b - c)]  = 

3(a - b)(b - c) (c - a)

OK, young person! I shall attempt to solve it for you but may charge you for it!! Just joking!.

First, you should be familiar with this complicated-looking formula, but in fact, it is just PV and FV formulas combined into one. If you know any 4 of the 5 variables, then you can solve for the fifth, except for interest rate!:

-P*[(1-(1+R)^-N)/(R)]+FV*(1+(R))^-N+PV=0, where P = Periodic payment, R= Interest Rate per period, N= Number of periods, FV = Future value, PV = Present value.
So, the first thing we have to do is figure out the original monthly payments for the two separate loans:

1 - From the first loan of $15,000, you enter the following values into the above formula: 6%/12 =0.005 under R, 60 under N, 0 under FV, 15,000 under PV. Then you would solve for P. If you don't make any mistakes, then you should get P = $289.99, which is the monthly on the $15,000 loan. Because he has already made 12 monthly payments, we have to find the balance of this particular loan. You would use the above formula again, but this time you would enter 12 under N and $289.99 under P and you would solve for FV, which would be the balance of the loan after 12 payments. If you don't make any mistakes, you should get: $12,347.98.

2 - For the second loan of $10,000, you would do exactly the same thing as above entering your numbers accurately. If you don't make any mistakes, you should get the monthly payment to be: $295.24 and the remaining balance after 12 payments to be: $6,798.86.

3 - Now, you must add the balance of the two loans and you get: $12,347.98 + $6,798.86 = $19,146.84 - which is the balance of the new consolidated loan!. You do the same with the two original monthly payments: $289.99 + $295.24 = $585.23 - which is the new combined monthly payment on the new consolidated loan for a term of 3 years.

4 - Finally, you have to enter the new loan of $19,146.84 under PV, $585.23 under P, 36 (3 x 12) under N, and 0 under FV, and you would solve for R, or the monthly interest rate. Except, unfortunately, you can't solve for R directly but you have to use iteration and interpolation(basically, trial and error) until you arrive at the right solution.

5 - My computer is programmed to solve such equations and it comes out with the rate of 6.3160745455% compounded monthly. So, the effective annual cost to him would be 6.502162744568%. And that is the END!.

Sorry to be dim, but can you explain why, for every number in a row to be even, k must only have even factors?

OK, Lightning....glad I could help   !!!

A polynomial is an expression in one  (ore more) variables

The  exponent on each variable must be a whole number  [ 0, 1, 2, 3....]

The polynomial must have a finite number of terms

Examples

6x^2  +  2x  +  8

(4/5)x^3y^5 + 2y^2  + 3x  -  9/13

The  red entries  are coefficients  [ real numbers associated with each variable term]

The blue  entries  are constants

Thanks Chris!

b )     rx + s

        _____   =  f(x)

        2x + t

If the horizontal asymptote  =  -4, then   r / 2  = -4 ⇒  r  = -8

If the vertical asymptote = 3, then  2(3) + t  = 0  ⇒  6 + t  = 0  ⇒  t   = -6

If  (1,0)  is an x intercept, then  r(1) + s  = 0 ⇒  -8(1) + s  = 0 ⇒ - 8 + s  = 0 ⇒

s  = 8

Here's a graph :  https://www.desmos.com/calculator/cown9bo626

Well done! Assuming you did that by hand it was quite an achievement! However, the fact that the question only asks for \(\alpha+\beta\) suggests that the questioner probably had a short cut in mind, like the one in my answer below.

A function f has a horizontal asymptote of y = -4, a vertical asymptote of x = 3, and an x-intercept at (1,0).   Find an expression for f(x)....

a)   ax + b

     ______  =  f(x)       

       x + c

If the vertical asymptote is 3, then , in the denominator,  3 + c  = 0  and  c   =  -3

If the hrizontla asymptote is -4, then  ax / x  = -4  ⇒  a   = -4

And if the x intercept = ( 1,0), this implies that   a(1) + b  = 0  ⇒  -4(1) + b  = 0  ⇒

-4 + b  = 0  ⇒   b  = 4

Here is a graph : https://www.desmos.com/calculator/ohx3qy9kvt

Since the left hand side is the ratio of two linear terms the right hand side must be too. In other words the numerator and denominator on the right hand side must contain a common factor, say \((x-\gamma)\) which cancels out. The right hand side can therefore be written:

\(\displaystyle{(x-\alpha)(x-\gamma)\over(x+\beta)(x-\gamma)} ={x^2-(\alpha+\gamma)x+\alpha\gamma\over x^2+(\beta-\gamma)x -\beta\gamma} ={x^2-80x+1551\over x^2+57x-2970}\),

so equating coefficients of x in the polynomials we must have \(\alpha+\gamma=80\) and \(\beta-\gamma=57\).

Adding these two expressions together \(\gamma\) cancels and we have immediately:

\(\alpha+\beta=80+57=137\).

Thanks for your interest. Sorry, I made a small mistake about the term of the new loan. It is actually for 3 years, NOT 4 as I stated in the question. I can give you the answer but prefer not to, so that you can come with your own original solution using right financial formulas and/or equations. Thanks again.

Yep! 

Coefficients are the numbers infront of a varible. A polynominal can only use addition, subtraction, multiplication (doesn't have to use all of them). The exponents of a polynominal cannot be negative. Normally they are in the form \(ax^2 + bx +c\) but doesn't have to be.

Examples: x^2 + 2x +5,

5x +1

(x^7 + 2x^4 - 5) * 3x

Would you mind giving me the answer so I could think through the logic?

Thanks! your right!

7.   

 y  = √[2x - 3]     square both sides

y^2  = 2x  - 3

y^2 + 3   = 2x     

[ y^2 + 3 ] / 2  =x

[x^2  + 3] / 2    = y

8. 

y = √[4x] - 7

y + 7  =√[4x]   square both sides

(y + 7)^2  = 4x

( y + 7)^2  / 4   = x

(x + 7)^2 / 4   = y

Is #7 this: \(y=\frac{x^2-3}{2}\)

Thank you, that's not quite how I learned it but maybe i'll get used to it being that way...

You are correct in saying 6 rows. Starting with row 0, the kth row has the numbers C(k, 0) C(k, 1) C(k, 2) ... C(k, k). For every number in the row to be even, k must only have even factors, so it must be a power of 2. Since the highest power of 2 under 100 is 2^6, 6 of the first 100 rows have only even numbers besides 1.