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Let a  be  the first  term

We have that

a + ar + ar^2   =  54

a ( 1 + r + r^2)  =  54

a =  54 / (1 +r + r^2)    ⇒   (1 + r + r^2)  =   54/a        (1)

And

54 + ar^3 + ar^4 + ar^5  =  70

ar^3 / (1 + r + r^2)    = 16  ⇒     (1 + r + r^2)  =  16/ (ar^3)      (2)

Equating (1) , (2)

54 / a  =  16 / (ar^3)         [  a   ≠  0 ]

54  =  16/r^3

r^3  =  16/54  =  8/ 27

r =   (8/27)^(1/3)   =  2 / 3

CORRECTION ....THX, CATMG  !!!

Let's work with the denominator

sqrt 2  +  1 /sqrt 8   =

sqrt 2  + 1 / [2sqrt 2 ]  =

[ 2sqrt 2 ]  * sqrt 2  + 1

________________    =

        2sqrt  2

2 * 2 + 1

_________   = 

   2sqrt 2

5 / sqrt 8

So

1 /  [ 5 / sqrt 8]  =

sqrt 8                  2sqrt 2

_____     or       _______     (simplified)

    5                          5

x^3 - 10x^2 > -25x

x^3 -10x^2 + 25x > 0

x^2 - 10x + 25 > 0

(x - 5)^2 > 0

The solution is all x except 5.  In interval notation, the soution is (-inf,5) U (5,inf).

Not great at these, but I'll give this a try

f(x) must  have  the point   (-6,3)   [ the inverse would reverse these coordinates  ] 

f (3x)  compresses  the graph  of f(x)  by  factor of 3  so ( -6/3 , 3) = ( -2,3)  would be on this graph

And   2 f(3x)   means  the point  (-2, 2*3)   =    (-2, 6)  is on this  graph

And  2 f(3x)  - 5    means the point  (-2 , 6 - 5)  =    ( -2, 1)   is on this graph

So

(1,-2)  is on  the inverse  of  g

M==Total muffins

Chocolate muffins==3/11M

Strawberry muffins==1  -  3/11 ==8/11M

3/11M  -  [2/3 * 3/11M] ==8/11M -  42

3/11M - 2/11M == 8/11M  -  42

1/11M ==8/11M - 42

1/11M - 8/11M == - 42

- 7/11M == - 42

M ==42 / (7/11) ==42 x 11 / 7==66 - total number of muffins she made

66  x  3/11 ==18 muffins were chocolates.

18  x  2/3 == 12 chocolates muffins were sold.

(a) - 66  -  18 == 48 muffins were strawberry.

[48  -  42] ==6 strawberry muffins left

[18  -  12] == 6 chocolate muffins left. 

[12  +  42]==54 muffins were sold

(b) - 54 / 66 == 9 / 11 - ratio of muffins sold to total muffins.

6mn  = 27 +m  - 2n

6mn +2n  =  27 + m

n ( 6m + 2)  = 27 + m

n  =   (27 + m)  / ( 6m  + 2)

If  m =  5   then   n =  1

(m , n)  = (5,1)

2x + y =  15    ⇒    y  =   15 -  2x       (1)

x + 2y  =  23           (2)

Sub   (1)  into (2)

x  + 2 ( 15 - 2x)  = 23

-3x + 30   =23

-3x  =  -7

x = 7/3    = the   value of one circle

So  2 circles  =  14/3

3^a = (3^4)^(b + 2)                 (5^3)^b = 5^(a - 3) * (5^2)^b       divide through by (5^2)^b

3^a  = 3^(4b + 8)                       5^b =  5^(a - 3)

a = 4b + 8                                      b = a - 3          (2)

a - 4b =  8       (1)

Sub  (2)  into (1)

a - 4 ( a - 3)   =8

a - 4a + 12  = 8

-3a  =  - 4

a = 4/3

And

b  = a  -3  =  4/3  - 3  =  -5/3

ab=  (4/3)(-5/3)  = -20  / 3 

\( $\displaystyle{ \frac{2}{1 + 2\sqrt{3}} + \frac{5}{2 - \sqrt{3}}}$\)

We have

2 [ 2 - sqrt 3 ] + 5 [ 1 + 2sqrt 3 ]

_________________________   =

  (1 + 2sqrt 3)  ( 2 - sqrt 3)

4 - 2sqrt 3  + 5 + 10sqrt 3

______________________  =

  2  + 4sqrt 3 - sqrt 3  - 2*3

9 + 8sqrt 3

_________   =     

-4 + 3sqrt 3

(9  + 8sqrt 3) ( -4 -3sqrt 3)

______________________  =

(-4 + 3sqrt 3) ( -4 - 3sqrt 3)

-36 - 32sqrt 3 - 27sqrt 3 - 24*3

__________________________  =

16 - 9*3

-108 - 59sqrt 3

___________  =

       -11

108 + 59sqrt 3

___________

         11

A  +  B  +  C  = 

108  + 59  + 11   = 

178

People here want feed back for their answers.

I am no exception.

Let the roots  be m and n

The sum of the roots  =  -5/a  =  m + n

Square both sides of this

25/a^2  = m^2  + 2mn + n^2

The product of the roots =  - 3/a = mn

Then    -4(-3)/a  =  -4mn

So      12/a  =  -4mn

So

25/a^2  + 12/a   =    m^2 + 2mn - 4mn  + n^2

25/a^2  + 12/a  =  m^2 - 2mn  + n^2

25/a^2  + 12/a  = (m +n) (m - n)

25/a^2 + 12/a  =  (-5/a)(m - n)        multiply through by   a

25/a  + 12  = -5 ( m - n)

If we let   m - n  =   -7/2     then we have that

25/a  + 12  = -5(-7/2)

25/a  + 12  =  35/2             multiply through by 2

50/a + 24  =  35

50 / a  =   21

a  =  50  / 21

See the follwing graph : https://www.desmos.com/calculator/ospph6s7lf

Note that  the line  y = -2x + 8   forms a right trangle with the axis....it has a height of 8  and a base of 4

Its area =    (1/2)(4)(8)  =  16

Likewise,  the line x + y  = 2      forms a right  triangle with the axis  and has a base  and height of 2

Its area  =  (1/2)(2)(2)  =  2

The polygon formed has an  area of  =   16  -  2   =  14

We have the following system:

\(6+b+c=24\)       (1)

\(36+b^2=c^2\)             (2)

Solve for \(c\) in the first equation: \(c=18-b\)

Squaring this gives \(b^2-36b+324\)

Substituting this in (2) gives: \(36+b^2=b^2-36b+324\)

Solving, we find that \(b= 8\)

Subsituting this in to (1), we find that \(\color{brown}\boxed{c = 10}\)

BG = 2*sqrt(3).

sorry thats not right apparently you reversed the action of the horizontal compression.

9 possibilities for the first and third digits, and if the second digit is 2, there are 6 possibilities for the last digit, and if the second digit is 4, there are 2 possibilities for the last digit.

So the answer is 9*9*(6 + 2) = 648.

The point is (-6,-13).

Solution: \(1\)

Answer:

We are looking for the number of sevens there are in the prime factorization of the number.

60! has 9 sevens in its prime factorization, because there are eight multiples of seven in 1x2x3x4...x60, and one of them is 7².

Because the 60! is being square, the number of sevens in its prime factorization doubles to 18.

120! has 19 sevens in its prime factorizations, because there are 17 multiples of seven in 1x2x3x4....x120, and two of them are multiples of 49.

\(x \cdot7^{18}\over y \cdot 7^{19}\)\(=x/y \cdot 7\)

where x and y are two numbers that aren't multiples of 7 (and so therefore neither is their quotient).

There is only one seven the prime factorization of this number, which means that the largest integer n such that 7^n that divides it is just \(1\).

The perfect squares below 1,000 are:

 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900 and 961

The answer is ...

Could you have figured out the answer without the list?

Translate  -3 <= a <= 2  into   -3 <= x <= 2  and  -1 <= b <= 4  into  -1 <= y <= 4 

and place this region onto a regular  x-y  plane.

The total region covered is  5 · 5  =  25  square units.

The positive product region is  (2 · 4) +  (-3 · -1)  =  8 + 3  =  11  square units.

The probability of landing on the positive region is  11 / 25.

See  https://web2.0calc.com/questions/help_54234

If you use the same calculations, you get an answer of 7/12.

Thanks for trying, however I am sorry to inform you that your answer is incorrect.

A will hold     pi (10)^2 *16  =  1600 pi  ft^3 of water when full

B will hold    pi (9)^2 * 18  =  1458 pi  ft^3  of water when full

A will hold   [  1600  - 1458 ]  pi   ft^3 of water after filling B   =  441.6 ft^3

You're  correct...it's impossible to answer this  !!!!!

Let a =  2n+ 1   and b  = 2n + 3

So

(2n + 3)^2  - (2n + 1)^2  = 48

4n^2  + 12n + 9  -  4n^2  - 4n - 1  = 48

8n  +  8  =  48       divide through by 8

n + 1  =   6

n =  5

So

a =  11      b =  13      c  = 15

And

c^2  -  b^2   =

15^2 -  13^2   =  56