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Hi @CPhill, thanks for all of your help. I was wondering if the answer to this question is Is it r=1/3, an=-24(1/3)^n-1?

Sum of a geometric series  =   a (1 - r^n) / (1 - r)

Where a is the first term,(1), r is the common ratio, (-3), and n is the number of terms summed

So  the sum of the first 5 terms  =

1[1 + (-3)^5] /(1 - 3)  = 61

We can make these equations from the info in the question:

a = s * s

3a = (s + 6) * (s + 3)

Substitute.

3(s * s) = (s + 6) * (s + 3)

3s² = s² + 9s + 18

Subtract 3s² from both sides.

0 = -2s² + 9s + 18

Now we can use the quadratic formula to solve for s.

\(s = {-9 \pm \sqrt{9^2-4(-2)(18)} \over 2(-2)} \\~\\ s= {-9 \pm \sqrt{81+144} \over -4} \\~\\ s= {-9 \pm \sqrt{225} \over -4} \\~\\ s= \frac{-9\pm15}{-4} \\~\\s =\frac{ -9 + 15} { -4}=-\frac32\text{ or } s=\frac{-9-15}{-4}=6\)

We know the side length is positive.

\(s=6\text{ cm}\)

A. 1st term   = 1460

B. We need to solve this

74  = 1460 + d(57 -1)

74   = 1460 + d(56)      subtract 1460 from both sides

-1386  =  56d      divide both sides by 56

d = -24.75

So the general equation is

a_n  = 1460 - 24.75(n - 1)

C. The 13th term is

1460 -24.75(13 -1)  =  1460 -24.75(12)  = 1163

We compare two surfaces.
The sum of the two surfaces is not interesting (given the given question).

write each expression as a single logarithm in simplest form. state any restrictions on the variable.

\(log_3\frac{\sqrt{x}}{x}+log_3\sqrt[5]{x^2}+\frac{3}{2}log_3x\)

\(=log_3\frac{\sqrt{x}\cdot \sqrt[5]{x^2}\cdot \sqrt{x^3}}{x}\)

\(=log_3(x^{\frac{1}{2}+\frac{2}{5}+\frac{3}{2}-1})\) 

\(=log_3(x^{1.4})\)

\(=1.4\times log_3(x)\)

  !

Not possible......to see why let x =  a/b  and let y  = c/d   where a,b,c,d   are integers

Then we have that

(a/b)^2   =  2(c/d)^2   rearrange as

(a/b)^2 /(c/d)^2   = 2

(a^2 *d^2)  /  (b^2 *c^2)   = 2         take the square root of both sides

(a * d) / (b * c)  =  √2

But  a*d  , b*c     are both integers.......let a*d  = m  and b*c  = n

m / n  = √2

But......the square root of two is irrational....so.....we're led to a false conclusion.....thus, x and y  are not both rational

We have that

S  = first term ( 1 - r^n) / (1 - r)  ......so.....

93   =  3  (1 - 2^n) ( 1 - 2)         divide both sides by 3

31  = (1- 2^n) / -1      multiply both sides by -1

-31  =  (1 - 2^n)        multiply both sides by -1 again

31  =  2^n  - 1      add 1 to both sides

32  = 2^n         →    n  = 5   terms

We have that the total vertical distance traveled  is :

              8

35 + 70 Σ .8^n     =   268.024 m  =  268 m 

              1

If the rate of reaction is 5 x 10-9 mol/s, how many grams of magnesium will be used up in 5 min?

\(\large 5\times10^{-9}\frac{mol}{s}\times\frac{24.305g\ Mg}{mol}\times 5min \times\frac{60s}{min}=3.64575\cdot 10^{-5}g\ Mg\)

  !

Hmm.. this is my (very limited!) understanding of probability...

 \(\text{probability}=\frac{\text{number of ways something can happen}}{\text{total number of possible outcomes}}\)

So... if you have a bag of 3 blue marbles and 8 red marbles,

the probability of drawing a blue marble = \(\frac{3}{11}\) (and not \(\frac38\))

How can I solve [(-4)5]9?

You can omit the staples.

\([(-4)5]9=-4\times5\times9 =-180\)

  !

@CPhill could you help me on this one as well?

@CPhill could you please help me?

Is it r=1/3, an=-24(1/3)^n-1

Oops.....thanks, hectictar, for catching my error....!!!

a(n) =20,000 x (1 +0.10)^16

a(n) =20,000 x    1.10^16

a(n) =20,000 x       4.5949729863572161

a(n) =$91,899.46

sorry it's log_na=5 and log_nb=3

The general formula is:  a_n  =  a₁ + (n - 1)d

a₈ = 28         --->     a₈  =  a₁ + (8 - 1)d        --->        28  =  a₁ + 7d

a₁₆ = 108     --->     a₁₆  =  a₁ + (16 - 1)d     --->     108  =  a₁ + 15d

Subtract down the columns:                                    -80  =        - 8d

                                                                        --->          d  =  10

Substituting this value for d into the first equation:  28  =  a₁ + 7d

                                                                         --->       28  =  a₁ + 7(10)

                                                                         --->       28  =  a₁ + 70

                                                                         --->        a₁  =  -42

So the equation is:  a_n  =  -42 + (n - 1)(10)

To find a₂₄, substitue 24 in for n and solve. 

The point X meets randomly from above on two surfaces of equal importance:

the circular section\((\leq 1 \ unit\ from\ vertex\ A)\) and the rest of the triangle\((>1 \ unit\ from\ vertex\ A)\).
The probability can only be the quotient of these two surfaces.

188    ...........

761.5314345041948334

log_n(a·b²)  when  a = 5  and  b = 3  is:  log_n(5·3²)  =  log_n(45)

Do you know the value of n?

I would estimate on 366.70 

If log₁₀c = 3, evaluate log₁₀c²

Since log₁₀c²  =  2·log₁₀c             (because the exponent in a log expression comes out as a multiplier)

--->    log₁₀c²  =  2·3   =  6