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5 root 13

Use the standard formula,

Vol = (1/3)*height*area of the base.

Calculate the area of the base as the sum of the areas of six equilateral triangles.

According to your equation:   2y = -5   so    y = -2.5

ALSO:     x + 2x - 3y = -5

               3x - 3(-2.5) = -5

               3x + 7.5      = -5

                3 x = -12.5

                   x = - 125/30 =  - 4 1/6

IF you entered your Q correctly !

Use this formula: S=n/2[2a + (n - 1)d], where S=sum of the series, n=number of terms, d=common difference between terms, a=first term.

what is x if ....

3x2 + 9x – 2 for x = -53

3*(-53)^2 + 9*(-53) - 2   [Is this what you wanted?]

8,427        + ( -477) - 2

8,427 - 477 - 2

=   7,948

Nice !

Still disagree.  At the beginning of the 7th year 'd' students have been added.     At the beginning of the 2nd year 'd' students have been added.       We'll just have to agree, as the English language is full of 'nuances' and unclear meaning, to disagree.  G'Day!    Thanx  !

35.72/180 x Pi =0.623432 Radians

The young lady says "At the beginning of its 7th year, it had 449 students". Which is the same as saying "At the end of the 6th year, it had 449 students"!!!

Let the amount of 8% bond =E

Let the amount of 5% bond =F

F+E=30,000

.08E - .05F =840, solve for E, F

E=$12,000

F=$18,000

Hey there! It's been a long time since I left, so it's gonna be my first answer in 3(or 4) months.

Anyway, back to the question.

    (9 - 3)x 1/3 + 1

= 6  2 x 1/3 + 1

= 2 + 1

= 3

I'll have to disagree with Guest #2.   

Your equations do not work if d=43  and a = 234

a(n) =a(1) + (n-1)d      results in a(7) = 492                 the correct answer needs to be 449

Your mistake is here:   449 =a(1) + (6-1)43      where '6' should be '7'      AND at year 7 the students are at 449  SIX (not SEVEN as you described)more terms later , at year 13, the total is 750.  

Soooooo the 'common difference' as you describe it is  (750-449)/6 = 50 1/6  (NOT 43).

Honest mistakes we all make (frequently!)    

Use this formula to find the amount of the annuity:

PV=P{[1 + R]^N - 1.[1 + R]^-N} R^-1

PV=200 {[1 + .05/12]^120-1 x [1+.05/12]^-120 x (.05/12)^-1

PV=200 x 94.281350

PV=18,856.27

Find the annual percentage rate (APR) for the following interest rates

(i) 5.39% compounded annually
(ii) 5.3% compounded semi-annually,

(iii) 5.2% compounded quarterly, (iv) 5.18% compounded monthly .

i) (1+0.0539/1)^1-1

apr = 5.39%....................CORRECT.

ii)5.4%............................5.370225%

iii)5.3% (0.05302)..........5.302282%

iv)5.3% (0.05304)..........5.304769%

cheapest is (iii)...............CORRECT.

Solve for x:
3 x^2+9 x-2 = 0

Divide both sides by 3:
x^2+3 x-2/3 = 0

Add 2/3 to both sides:
x^2+3 x = 2/3

Add 9/4 to both sides:
x^2+3 x+9/4 = 35/12

Write the left hand side as a square:
(x+3/2)^2 = 35/12

Take the square root of both sides:
x+3/2 = sqrt(35/3)/2 or x+3/2 = -sqrt(35/3)/2

Subtract 3/2 from both sides:
x = sqrt(35/3)/2-3/2 or x+3/2 = -sqrt(35/3)/2

Subtract 3/2 from both sides:
Answer: | x = sqrt(35/3)/2-3/2    or     x = -3/2-sqrt(35/3)/2

The difference between the 13th term and the 6th term is:

13 - 6 =7 terms

The difference between the values of the two terms is:

750 - 449 =301

Therefore, the common difference between the two terms is:

301/7=43

a(n) =a(1) + (n-1)d...................(1)

449 =a(1) + (6-1)43

449=a(1) + 215

a(1) =449 -215

a(1) =234

a(n) =a(1) + (n - 1)d.................(2)

750=a(1) + (13- 1)43

a(1) =750 - 516

a(1) =234

N5 + N8 = 30000

0.08*N8 - 0.05*N5 = 840. Replace N8 using the above:

0.08*(30000 - N5)  - 0.05*N5 = 840

0.13*N5 = 30000 - 840

N5 = (30000 - 849)/0.13

N5 = $12000

18,000? cuz that's not the answer.... That may be how much he invested in the 8% bonds. We're trying to figure out how much he invested in the 5% bonds.

Oh yeah, the answer is 12,000. Thanks though!

x = amount at 8%

30,000 - x  = amount at 5%

x(.08) - 840 = (30,000 - x)(.05)     Solve for 'x' and '30,000 - x'

@@ What is Happening?  [Wrap4]   Thurs 28/7/16   Sydney,  Australia Time 10:43am   ♪ ♫

Good day to you all,

I have not done a wrap since Sat 7th May 16, and that was the first one since Feb 16.

I stopped doing them because the forum got too big and I was too busy with other things.

At this stage I do not intend to start them again on a remotely regular basis but there were a couple of interesting post today and I just wanted people to see them.

I am sorry, I know many of you post and answer very interesting questions very often and I do not even see them anymore.  I am sad about this.  If you send them to me by private message and I have time I will post them in a wrap for you so that other people can see them too.

Forum  News:

I was just reading my last news entry and the problem is the same.  Most people cannot edit their posts!  This is a real pest and it is one of the reasons that I stopped writing the wrap.  Collecting the questions just became too difficult when i could not edit the thread that  they were entered into.  Soon I will apprach Mr Massow about this again but being realistic the problem is not likely to go away.

Camelot News:

Camelot seems to have been defeated.   

The King is still off at the crusades.  Queen Guinevere has not been doing a good job of ruling in his stead.  The ghost has been silent.  The Doctor, Morgan Tudd, has not been seen in months.  Sir CPhill and Sir Alan are here but they seem to be in disguise. I have heard no more news of Sir CPhill's search for the Roman Zero.  He knows it is in Sisyphus's boulder but still he has not retrieved it.    I did see Mistress Katie Leigh recently but she was vas very quiet. I did not even hear her school keys jangling.  Maybe our reporter HeatherF, is off soing war correspondence with the King.  I have not seen her.    

Mathematics posts:  

I am sure that there has been a great many excellent questions and answers but unfortunately I have not seen them. :(

1)  Max posted this one and Heureka and I answered it.  Thanks Heureka :)

a = #students first year

a + d = #students SECOND year

for all of the following 'n' years

#students = a + (n-1) d

For 7   years  #students = 449 = a + (6)d            (1)

For 13 years  #students = 750 = a + (12)d          (2)

From (1)     a = 449 - 6d 

Substitute into (2):      750 = (449 - 6d) + 12d  and solve for 'd'

                                    750-449 = -6d + 12d

                                     301 = 6 d  

                                       d=50 1/6  

     (not sure how you add 50 1/6 students per year...must be an AVERAGE per year)

Substitute this value of 'd' into one of the equations to find 'a' = ?

(1)    449 = a + 6 (50 1/6)

         449 = a + 301

         449 - 301 = a             a = 148