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Just look at the hoghest powers

2x^4 * 3x^10 = 6x^14

6x^14 ..... -   x^16 .....

Highest power of x will be 16     The leading term will be    -x^16

the degree is 16

a + c + e - (b + d + f) = 23.

Just sum it up as an infinite sequence: 

First term=1/7 + 2/7^2 + 3/7^3 =66 / 343

Common Ratio =1 / 343

Sum=(66/343) / (1 - 1/343) =11 / 57 - the answer!

\[\sin^2 θ = 1 - \cos^2 θ = 1 - \frac{4}{13} = \frac{9}{13}\]

every angle in a parallelogram is between 0 and π radians, so θ must be positive, so

\[\sin θ = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}\]

area of parallelogram is

\[\|{a}\| \|{b}\| \sin \theta = \sqrt{65} \cdot \sqrt{5} \cdot \frac{3}{\sqrt{13}} = \boxed{15}\]

 ∑[ 66 / 343] * [1 / 343]^n, n, 1, ∞]==It converges to 11 / 19551

a = 21/5.

a = 9 and b = 8, so a + b = 17.

FORGET THE ABOVE!  IT IS WRONG.

The answer is 17.

sumfor(n, 1, infinity, (n / (7^n))==converges to 7 / 36

let's call the money he initally had = $x

when he removes $12 from $x he had 3/7 of $x left

$x - $12 = 3/7 of $x

($x - $12) = $3x/7

7 ($x - 12) = $3x

$7x - $84 = $3x 

$7x - $3x = $84

$4x = $84

$x = 21

Therefore he initially had $21.

$21 - (how much spent) = (what's left)

$21 - $12 = $9.

$9 was left

Let Brian's initial amount be x

x - 8 = 2/3x

3 (x - 8) = 2x 

3x - 24 = 2x

3x - 2x = 24

x = 24

Amount left = 2/3x = 2/3 * 24

                   = 16

Brian has $16

1/x  + 1/y =  3

1/x  -  1/y  = 7          add these

2 / x =  10

x = 2/10  = 1/5

And  

1 / (1/5) + 1/y = 3

5 + 1/y=  3

1/y =  3 -5

1/y =  -2

y  = -1/2

x + y =   1/5  - 1/ 2 =   -3 /  10

a@b  =  ab - 2b                   a#b = a + b  - ab

So

(6 @ 2) /   (6 # 2)  =

(6*2 - 2*2) / ( 6 + 2  - 6*2)  =

(8 ) / ( -4)  =

-2

Getting a common  denominator we have

30/60      20/60       15/60     4/60

We need the result to be  < 3/60  (but positive)

We have 8 possibilities

 + + +           reject  as too large

+ + -            reject  as too large

+ - -             reject as too large

 - -  -              reject  (negative result)

- - +               reject  (negative result)

- + +              reject as too large

+ - +            reject as too large

- + -              reject as too large

The probability is  0

M - 12 = (3/7)M

M - (3/7)M =  12

(4/7)M = 12

M = 12 (7/4) =  $21   is what he started with

He has   21 - 12 = $9  left

Let the number that Tom had = T

Then Peter had 4T

When tom got 48  pencils, Tom only had twice as many

So

2 *(T + 48)  = 4T

2T + 96   = 4T

96 =  4T - 2T

96 = 2T

96/2  = T =  48

a )   Tom had   48 + 48  = 96 at the end

b )  Peter had  4 (48) =  192

Together, they had     96 + 192 =  288
credit to cphill

Let \(x = \sqrt{-16 + 31i}\) and be in the form a + bi.

We have: \((a+bi)^2 = (a+bi)(a+bi) = a^2 + 2abi - b^2 \)

We know that \(2abi = 31\), meaning \(ab = {31 \over 2}\)

Now, remember that \(a^2 - b^2 = -16\)

This gives us the system: 

\(a^2 - b^2 = -16\)          (i)

\(ab = {31 \over 2}\)                       (ii)

Using WA, we find the answer to be: \(\sqrt{-16 + 31i} = \color{brown}\boxed{\sqrt{{\sqrt{1217}\over 2} - 8} + \sqrt{8 + {\sqrt{1217}\over 2}}i}\)

Let the number that Tom had = T

Then Peter had 4T

When tom got 48  pencils, Tom only had twice as many

So

2 *(T + 48)  = 4T

2T + 96   = 4T

96 =  4T - 2T

96 = 2T

96/2  = T =  48

a )   Tom had   48 + 48  = 96 at the end

b )  Peter had  4 (48) =  192

Together, they had     96 + 192 =  288

let's write the equation 

5x/6 + x/10 = x/10 + 50

5 * x/6 = 50

5x/6 = 50

5x/6 * 6 = 50 * 6

5x = 300

x = 60

Let  Hunter's seashells be H

Then Adam has 2H

When  Hunter finds  15 more we have that he now has twice as many as Adam....so....

H + 15  =  2 ( 2H)    simplify

H + 15  = 4H

15 = 4H  -H

15 = 3H

15 / 3 =  H =  5

Adam had  2H =   2 (5)  =  10 shells

6/5 * 50 = 60

so he actually has 60 pennies