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S1 = 5

S2 = 15

Then the second term is 10   and  d = 5

15th term  is  5 + 14(5) =  75

S15  = [ 15th term + first term ] * no.of terms /  2 =

 [ 75 + 5 ] (15 / 2)   = 80 * 15/2  =  40 * 15  = 600

Hi Chris, 

So, I blundered into the right answer but for the wrong reason. 

All the wrong reasons.  Thanks for posting the correct answer. 

Ron  

_.

a1 + a3 =  12

a1  +  (a1 + 2d)  = 12

2a1 + 2d  =  12

a1 + d =  6

d = 6 - a1

a2 + a4 + a6  = 8

(a1 + d) + (a1 + 3d) + (a1 + 5d)  = 8

3a1 + 9d  =  8

3a1 + 9(6 - a1)  =  8

3a1 + 54 - 9a1  = 8

-6a1  = -46

a1 = -46 / -6  =  23 / 3

Simplify as 

x^2 - mx + 14 =  0

Roots             m

-1  -14            15

1    14           - 15

-2   -7               9

2     7              -9 

What are the coordinates of the points where the graphs  of f(x)=x^3 + x^2 - 3x + 5 and g(x) = x^3 + 2x^2 intersect?

Hello ABJelly!

\(x^3 + x^2 - 3x + 5 = x^3 + 2x^2\\ x^2+3x-5=0\\ x_{f,g}=-1.5\pm\sqrt{2.25+5}\\ x_{f,g}=\frac{1}{2}(3\pm\sqrt{29})\\ x_{f,g}\in \{ -4.19258,\ 1.19258\}\)

\(y_g\in \{-38.540659,4.540659\}\\ \color{blue}So\ ( -4.19258,-38.540659),(1.19258,4.540659)\)

 !

n + n + d + n + 2d  = 180

3n + 3d  =  180

n + d =  60

n = 60  - d

( n + 2d)  - n =  56

2d = 56

d = 28

n = 60 - 28 =  32°

sqrt [ 4x + 1]   not well defined whenever  x < -1/4           x ≥   -1/4

sqrt [ 8 -3x ]   not well defined when x > 8/3                     x ≤  8/3

sqrt [ 2x - 5 ]  in the  denominator  not well defined when x ≤  5/2              x > 5/2

The most restrictive interval is  (5/2 , 8/3 ]   { well-defined here }

Note....either  l a  l  =  30       or   l -a  l  =  30

x^2 - 30x - 1 =  30

x^2 -30x - 31  =  0

(x -31) ( x + 1)  = 0

x = 31   and x =  -1

Also

- (x^2 - 30x - 1)   = 30

-x^2 + 30x + 1 = 30

x^2 - 30x - 1 = -30

x^2 - 30x + 29 =  0

( x - 1) ( x -29)  = 0

x  = 1  and  x =  29

x^2 - x - 2 =  (x - 2) ( x + 1)

Multiply through by the factorization and we get

2x + 11 =  A(x + 1)  + B( x  -2)

2x + 11 = (A + B)x + (A - 2B)

Equating terms

A + B  = 2     →   B = 2 - A    (1)

A - 2B  = 11     (2)

Sub (1)  into (2)

A - 2 ( 2 -A)  = 11

A - 4 + 2A  = 11

3A = 15

A = 5

B = 2 - 5  =  -3

Law of Cosines

BC^2  = BA^2  + AC^2  -2 ( BA * AC) cos (DAE)

18^2 = 10^2 + 14^2  - 2 ( 10 * 14)  cos (DAE)

[ 18^2 - 10^2 - 14^2 ] / [ - 2 (10 * 14) ] =  cos (DAE)

-1/10  = cos (DAE)

Again

DE^2  =  AD^2 + AE^2 - 2(AD * AE) cos (DAE)

DE^2  = 28^2 + 20^2 - 2(28 * 20) (-1/10)

DE^2  = 1296

DE  =  sqrt [1296]    = 36

                               A

                          1

                     F

                 2

           D                                     E

B                                                             C

FE patallel to DC so triangles AFE  and ADC  are similar

AF / FE  = AD / DC

1 / FE  = 3 / DC

FE = (1/3)DC

3FE = DC

FE parallel to DC   and DE parallel to BC  so triangle FDE  is similar to triangle DBC

FD / FE = BD /  DC

FD / FE = BD /  (3FE)

FD = BD /  3

3FD = BD

3*2 = BD =  6 

1. Call the center of the circle O

[AOP ] = (1/2) (2)(4)  =  4

OP =  sqrt [ 4^2 + 2^2] = sqrt [ 20 ] = 2sqrt 5

Height of triangle AOP can  be found as

4 = (1/2) 2 sqrt 5  *  h

h =  4/sqrt 5  =   AB /  2

AB  =   2 * 4/sqrt 5  =  8 / sqrt 5

3.

Angle XTA = 180 - 47 =  133

Angle XUA = 180  - 65  = 115

Minor arc TA =  2*47 =  94

Minor arc UA = 2*65 = 130

Arc TU =  360 - 94 - 130 =  136

Angle TAU =  136 / 2 =  68

Angle QXR  =  360  - 133  - 115  -  68    = 44

Side of the square  = [ product of the legs ] / [ sum of the legs ]  =  

[ 6 * 6 ]  / [ 6 + 6 ] = 

36 / 12   = 

3

sorry, 144 is incorrect, as well as akhains answer

Friend 1        Friend 2

11                    1

10                    2

.

.

.

1                   11

11  ways

Since it's not specified whether Magnus can have stickers left over:

This is the solution if you don't have to hand out every sticker.

a + b <= 12.

Create a dummy variable, k, for all leftovers

a + b + k = 12.

This is great because we see k has to be at least 0, like all the others.

Apply stars and bars formula:

\({12+3-1\choose 3-1}=91\).

There are 91 ways.

This is the solution id you don't have to hand out every sticker.

Give one friend any number of stickers 0 -12. 

The other gets all the rest.

There are 13 ways.

f(1) =  -1

f(2) =   0

All   the other terms evaluate to 0   because the inside function will approach (but never reach) y = 1 since this is the horizontal  asymptote of that  function

So.....the sum  =  -1

https://web2.0calc.com/questions/rates_56

https://web2.0calc.com/questions/need-help-with-sum

The first one

We can  tell if it  has a repeated root if  the  discriminant =  0 

So

(-2)^2  -4(1) (1)  = 

4 - 4 =  

0

9  ( G - 7)  =  D - 7    →  9G - 63  = D - 7  →  9G = D + 56     (1) 

6 ( G - 4)  = D - 4     →  6G  - 24  = D - 4  →  6G  =  D + 20    (2)

Multiply   (2)  by 1.5

9G = 1.5D + 30     (3)

Set ( 1)  = (3)

1.5D  + 30  =  D + 56

.5D  = 26

(1/2)D =  26               mult both sides by 2

D =  52

Simplify as

x^2 + 6x + 5 =  0             factor as

(x + 3)(x + 2)  = 0

The roots are   -3 and  -2

(-3)^3 / (-2)^3   +  (-2)^3 / (-3)^3   =

27 / 8  + 8 / 27   =

793 /  216