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8==F * R^2,   

64==F *  R^5, solve for F, R

Divide the 2nd by the 1st and should get:

F==2 - this is the first term

D==2 - this is the common ratio.

9th term ==[2 * 2^8]==512

Your sequence looks like this:

2 , 4 , 8 , 16 , 32 , 64 , 128 , 256 , 512 , 1024 , >>Number of terms = 10>>Total Sum == 2046

1,  2, 2,  3,  3,  4 ==6 digits ==6! /2!.2! ==180 permutations as follows:

[(1, 2, 2, 3, 3, 4), (1, 2, 2, 3, 4, 3), (1, 2, 2, 4, 3, 3), (1, 2, 3, 2, 3, 4), (1, 2, 3, 2, 4, 3), (1, 2, 3, 3, 2, 4), (1, 2, 3, 3, 4, 2), (1, 2, 3, 4, 2, 3), (1, 2, 3, 4, 3, 2), (1, 2, 4, 2, 3, 3), (1, 2, 4, 3, 2, 3), (1, 2, 4, 3, 3, 2), (1, 3, 2, 2, 3, 4), (1, 3, 2, 2, 4, 3), (1, 3, 2, 3, 2, 4), (1, 3, 2, 3, 4, 2), (1, 3, 2, 4, 2, 3), (1, 3, 2, 4, 3, 2), (1, 3, 3, 2, 2, 4), (1, 3, 3, 2, 4, 2), (1, 3, 3, 4, 2, 2), (1, 3, 4, 2, 2, 3), (1, 3, 4, 2, 3, 2), (1, 3, 4, 3, 2, 2), (1, 4, 2, 2, 3, 3), (1, 4, 2, 3, 2, 3), (1, 4, 2, 3, 3, 2), (1, 4, 3, 2, 2, 3), (1, 4, 3, 2, 3, 2), (1, 4, 3, 3, 2, 2), (2, 1, 2, 3, 3, 4), (2, 1, 2, 3, 4, 3), (2, 1, 2, 4, 3, 3), (2, 1, 3, 2, 3, 4), (2, 1, 3, 2, 4, 3), (2, 1, 3, 3, 2, 4), (2, 1, 3, 3, 4, 2), (2, 1, 3, 4, 2, 3), (2, 1, 3, 4, 3, 2), (2, 1, 4, 2, 3, 3), (2, 1, 4, 3, 2, 3), (2, 1, 4, 3, 3, 2), (2, 2, 1, 3, 3, 4), (2, 2, 1, 3, 4, 3), (2, 2, 1, 4, 3, 3), (2, 2, 3, 1, 3, 4), (2, 2, 3, 1, 4, 3), (2, 2, 3, 3, 1, 4), (2, 2, 3, 3, 4, 1), (2, 2, 3, 4, 1, 3), (2, 2, 3, 4, 3, 1), (2, 2, 4, 1, 3, 3), (2, 2, 4, 3, 1, 3), (2, 2, 4, 3, 3, 1), (2, 3, 1, 2, 3, 4), (2, 3, 1, 2, 4, 3), (2, 3, 1, 3, 2, 4), (2, 3, 1, 3, 4, 2), (2, 3, 1, 4, 2, 3), (2, 3, 1, 4, 3, 2), (2, 3, 2, 1, 3, 4), (2, 3, 2, 1, 4, 3), (2, 3, 2, 3, 1, 4), (2, 3, 2, 3, 4, 1), (2, 3, 2, 4, 1, 3), (2, 3, 2, 4, 3, 1), (2, 3, 3, 1, 2, 4), (2, 3, 3, 1, 4, 2), (2, 3, 3, 2, 1, 4), (2, 3, 3, 2, 4, 1), (2, 3, 3, 4, 1, 2), (2, 3, 3, 4, 2, 1), (2, 3, 4, 1, 2, 3), (2, 3, 4, 1, 3, 2), (2, 3, 4, 2, 1, 3), (2, 3, 4, 2, 3, 1), (2, 3, 4, 3, 1, 2), (2, 3, 4, 3, 2, 1), (2, 4, 1, 2, 3, 3), (2, 4, 1, 3, 2, 3), (2, 4, 1, 3, 3, 2), (2, 4, 2, 1, 3, 3), (2, 4, 2, 3, 1, 3), (2, 4, 2, 3, 3, 1), (2, 4, 3, 1, 2, 3), (2, 4, 3, 1, 3, 2), (2, 4, 3, 2, 1, 3), (2, 4, 3, 2, 3, 1), (2, 4, 3, 3, 1, 2), (2, 4, 3, 3, 2, 1), (3, 1, 2, 2, 3, 4), (3, 1, 2, 2, 4, 3), (3, 1, 2, 3, 2, 4), (3, 1, 2, 3, 4, 2), (3, 1, 2, 4, 2, 3), (3, 1, 2, 4, 3, 2), (3, 1, 3, 2, 2, 4), (3, 1, 3, 2, 4, 2), (3, 1, 3, 4, 2, 2), (3, 1, 4, 2, 2, 3), (3, 1, 4, 2, 3, 2), (3, 1, 4, 3, 2, 2), (3, 2, 1, 2, 3, 4), (3, 2, 1, 2, 4, 3), (3, 2, 1, 3, 2, 4), (3, 2, 1, 3, 4, 2), (3, 2, 1, 4, 2, 3), (3, 2, 1, 4, 3, 2), (3, 2, 2, 1, 3, 4), (3, 2, 2, 1, 4, 3), (3, 2, 2, 3, 1, 4), (3, 2, 2, 3, 4, 1), (3, 2, 2, 4, 1, 3), (3, 2, 2, 4, 3, 1), (3, 2, 3, 1, 2, 4), (3, 2, 3, 1, 4, 2), (3, 2, 3, 2, 1, 4), (3, 2, 3, 2, 4, 1), (3, 2, 3, 4, 1, 2), (3, 2, 3, 4, 2, 1), (3, 2, 4, 1, 2, 3), (3, 2, 4, 1, 3, 2), (3, 2, 4, 2, 1, 3), (3, 2, 4, 2, 3, 1), (3, 2, 4, 3, 1, 2), (3, 2, 4, 3, 2, 1), (3, 3, 1, 2, 2, 4), (3, 3, 1, 2, 4, 2), (3, 3, 1, 4, 2, 2), (3, 3, 2, 1, 2, 4), (3, 3, 2, 1, 4, 2), (3, 3, 2, 2, 1, 4), (3, 3, 2, 2, 4, 1), (3, 3, 2, 4, 1, 2), (3, 3, 2, 4, 2, 1), (3, 3, 4, 1, 2, 2), (3, 3, 4, 2, 1, 2), (3, 3, 4, 2, 2, 1), (3, 4, 1, 2, 2, 3), (3, 4, 1, 2, 3, 2), (3, 4, 1, 3, 2, 2), (3, 4, 2, 1, 2, 3), (3, 4, 2, 1, 3, 2), (3, 4, 2, 2, 1, 3), (3, 4, 2, 2, 3, 1), (3, 4, 2, 3, 1, 2), (3, 4, 2, 3, 2, 1), (3, 4, 3, 1, 2, 2), (3, 4, 3, 2, 1, 2), (3, 4, 3, 2, 2, 1), (4, 1, 2, 2, 3, 3), (4, 1, 2, 3, 2, 3), (4, 1, 2, 3, 3, 2), (4, 1, 3, 2, 2, 3), (4, 1, 3, 2, 3, 2), (4, 1, 3, 3, 2, 2), (4, 2, 1, 2, 3, 3), (4, 2, 1, 3, 2, 3), (4, 2, 1, 3, 3, 2), (4, 2, 2, 1, 3, 3), (4, 2, 2, 3, 1, 3), (4, 2, 2, 3, 3, 1), (4, 2, 3, 1, 2, 3), (4, 2, 3, 1, 3, 2), (4, 2, 3, 2, 1, 3), (4, 2, 3, 2, 3, 1), (4, 2, 3, 3, 1, 2), (4, 2, 3, 3, 2, 1), (4, 3, 1, 2, 2, 3), (4, 3, 1, 2, 3, 2), (4, 3, 1, 3, 2, 2), (4, 3, 2, 1, 2, 3), (4, 3, 2, 1, 3, 2), (4, 3, 2, 2, 1, 3), (4, 3, 2, 2, 3, 1), (4, 3, 2, 3, 1, 2), (4, 3, 2, 3, 2, 1), (4, 3, 3, 1, 2, 2), (4, 3, 3, 2, 1, 2), (4, 3, 3, 2, 2, 1)] >Total distinct permutations = 180

The angles in a right triangle form an arithmetic progression.  If the smallest angle is 13 degrees, then what is the largest angle?  

It's me again.  I just re-read the question.  I overlooked that it says right triangle.  Disregard answer #1.  

So one angle of a right triangle is 90 degrees.

Since the other two angles have to total the other 90 degrees, each of them is less than 90 degrees.  

Therefore the largest angle is 90 degrees.  

BTW, it's not necessary to determine the arithmetic progression, if there is one. 

_.

The angles in a right triangle form an arithmetic progression.  If the smallest angle is 13 degrees, then what is the largest angle?     

An arithmetic progression means you get them by addition.  So,  

Let the 1st angle be called (13)  

Let the 2nd angle be called (13 + D)    D for difference  

Let the 3rd  angle be called (13 + D + D) 

We know that the sum of all three angles of a triangle is 180  

(13) + (13 + D) + (13 + D + D)  =  180  

39 + 3D  =  180 

        3D  =  180 – 39  =  141  

          D  =  141 / 3  =  47  

Thus  

The 1st angle  =  13 

The 2nd angle  =  (13 + 47)  =  60 

The 3rd angle  =  (13 + 47 + 47)  =  107 degrees

Check:  13 + 60 + 107 = 180  

_.

Any more info?

The height of a right cylinder is 2.5 times its radius. If the surface area of the cylinder is 60 cm^2,  what is the radius of the cylinder in centimeters?   

Does this cylinder have end caps, and if so are they flat or hemisphericalt; or are the ends of the cylinder open?  I'm not going to address every possible configuration.    

x=3, y=-2

the answer is 5

The answer is 5 when  x=3, y =-2

By long division, (-3x^2 + 12x + 22)/(x - 5) = -4x -4 + 9/(x - 5) = 9/(x - 5) + (-4)(x + 1), so A + B = 9 + (-4) = 5.

$f^{-1}(-3) + f^{-1}(0) + f^{-1}(3) = 3 + 1 + 4 = 8$.

By the quadratic formula, $x = \frac{-4 \pm 4 \sqrt{13}}{7}$, so $\frac{acd}{b} = \frac{(-4)(13)(7)}{4} = -91$.

Taking the derivative with respect to x and y, we get that the minimum value occurs at x = 6 and y = -4.  This gives us a minimum value of 6^2 + (-4)^2 - 6*6 + 4*(-4) + 18.

7 the ceiling was 4 the floor was 3

 5^3*25^6*3^3*4^7 ==13,500,000,000,000,000

The product has 14 zeros.

The point is (1,3,-4).

94 ==F + 26D,   

46==F + 93D, solve for D, F

Use substitution to get:

F==7546 / 67 - this is the first term

D== - 48 / 67 - this is the common diffrence.

Your sequence looks like this:

7546 / 67 , 7498 / 67 , 7450 / 67 , 7402 / 67 , 7354 / 67 , 7306 / 67 , 7258 / 67 , 7210 / 67 , 7162 / 67 ...etc.

BC = 12, AC = 14, the area of triangle ABC is 42.

1, 2, 3, 4, 5, 6, 7, 8, 9      O==Odd,   E==Even

O, E, O, E, O, E, O, E, O

There are 5 odd numbers ==5! 

There are 4 even numbers =4!

Total number of arrangements ==5! x 4! ==2,880

says the answer was 0

it says its wrong =(

There are six integers that work: -2, -1, 0, 1, 2, 3.

Let the number ==N

4N  +  20 ==139  -  3N

4N  +  3N ==139  -  20

7N ==119

N ==119 / 7 ==17 - which the number in question.