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First term:  a

Second term:  a + d

Third term:  a + 2d

Eleventh term:  a + 10d

Third term is the geometric mean of the first term and the eleventh term:

     a + 2d  =  sqrt( a · (a + 10d) )     --->     (a + 2d)²  =  a(a + 10d)

                                                         a² + 4ad + 4d²  =  a² + 10ad

                                                                 4ad + 4d²  =  10ad

                                                                           4d²  =  6ad

                                                                             2d  =  3a

                                                                               d  =  (3/2)a

Ratio of the second term to the first term: 

     (a + d) / a     --->   ( a + (3/2)a ) / a  

                                       (2a + 3a) / (2a)            (mult num and den by 2)

                                                 5a / 2a

                                                    5/2

sumfor(k, 1, 1000, (k / (3^k)))= it converges to 3/4.

Alright, thanks!  I will try what you said :)

Before you look at answer: Try to divide using long division... watch this video for more information: https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-long-division-of-polynomials/v/dividing-polynomials-1

Tried it already? Now, here is how I did it:

 What can you do to 12y^3-6y^2 to make it easier to divide? (*cough* factoring out a 6? *cough*) 

So, after I factored out a 6, I got 6*(2y^3-y^2)/(2y-1). This was easy to divide, and I got 6*y^2, which is 6y^2

OR

If you were smart enough to see if right after factoring, you could multiply the bottom by 6. This is 12y-6. Now, the difference between the top and the bottom? y^2. So, you still end up with 6y^2

:)

Since you already proved that n(2n + 1)(7n + 1) is divisible by 3, now you need to see how it could be divisible by 4. We have already seen that 2n+1 is odd, so n(7n+1) has no be divisible by 4. This happens when n is divisible by 4, or when 7n+1 is divisible by 4. Now, we first need to see when 7n+1 is divisible by 4. As you see, this happens if and only if -n-1 is divisible by 4. (by subtracting 8n) From there, can you generalize an equation that is true for n? (In the problem, you do not have to list out e v e r y   s i n g l e option, as there are infinite possibilities.)

(please ask for more help if needed!)

:)

Here is one of the many ways you can approach this problem:

As you see, there is a pattern: eachtime, you are taking away a digit, starting from 0.

So, we have

6075, 2025, 675, 225, 75, 25 (see the pattern? the last 2 digits are always the same)

Using this pattern, we see that for the last 2 digits to remain the same, it cannot go any further without becoming a fraction. So, there are a total of 6 integers.

We first need to find the prime factorization of 540. This is 2 × 2 × 3 × 5 × 3 × 3, which is 2^2(3^3)(5^1).

I never solved a system of equations by graphing before, but let me try!

Ok let's do this

the diameter of circle  M is [36] so the radius is 18, the area of semicircle M is 18^2*pi which is 324pi then /2 which is 162pi

Now let's move on, the radius of circle O=32+36 which is 68, that will equal to (68^2)/2 which is 2312 and the diameter is [136]

Now with these two Diameters, we can find the area of circle K, first we find the radius, that will be (136-36)/2 which is 50, then 50^2/2 which is 1250, finally multiply by pi and we get

                                                                                           1250pi

What I did (and kind of adding on to geno's answer):

We know that the median is 7. So, we can put 7 down as d.

We also see that for the mode to be 7, it has to occur the most. The most it can occur is 3, because if it occured less, than there would be 2 modes, and if it occured more, than g would not be that large. 

Now, for the mean to be 7, the a+b+c+7+7+7+g has to equal 49.

Subtract 21 from 49, and you get 28. From 28, you should see that to get the largest possible value of g, you need the smallest possible values of a, b, and c. Since the mode has to be 7, you can only have 2 1's. That leaves you over with one 2.

Subtract all of that from 28, and you get 24 for g.

:)

Try dividing the shape up into triangles.

You should end up with 4 triangles: ECD, CFE, BFC, and ABF.

Now, use the trangle area formula and some pythagorean (to find the heights) to get the area of each of those triangles. Then, add it up.

:)

if 8^x = 27, then what is 4^(2x-3)?

Hello Leaded!

\(x=\frac{log(27)}{log(8)}\)

\(4^{2x-3}=4^{2\cdot \frac{log(27)}{log(8)}-3}=\color{blue}1.265625=1+\frac{17}{64}\)

  !

Wrong

If you are driving at 60 km/h, that means you will cover 60 km in 1 hour. But if you cover that same distance in 45 minutes, which is =45 / 60 =3 /4 of an hour, then your speed will be:

60 / (3/4) = 60 x 4/3 =240 / 3 = 80 km/h - which is the answer.

If the sum of all digits of all the pages of the book sum up to 6869, then that means the book must have:

1,994 pages. And 1,994 =[1 + 9 + 9 + 4] =23 - the sum of the digits of N.

sine PQR  =  opposite / hypotenuse

sine PQR  =  12 / 13  

sine PQR  =  0.92307   

sine^–1 (0.92307)  =  67.38^o  =  correct to 1 decimal place is 67.4^o     

_.

See the answer here:  https://web2.0calc.com/questions/not-sure-how-to-solve-this_2

This graph should help:

tan theta + tan phi = 5/2 - 4/3 = 7/6.

The sum of the possible values of x is (-13) + (-3) = -16.

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My guess:  1, 1, 2, 7, 7, 7, 24   --->   24

I will try to find the white area by adding together the areas of the circles and subtracting out the area of the overlap.

The area of each circle is  pi·6²  =  36pi; so the area of the two circles is  72pi.

Now, for the overlap:

Call the center of the left-hand circle L, the center of the right-hand circle R, the top point of intersection T, and the 

bottom point of intersection B.

Since LT = 6, RT = 6, and LR = 6, angle(LRT) = 60^o and angle(BRT) = 120^o.

To find the segment bounded by arc(TLB) and segment(TB):

area  =  sector(TRB) - triangle(TRB)  =  (120^o/360^o)·36pi - ½·6·6·sin(120^o)

                                                            =           12pi           -    9·sqrt(3)

area of both segments  =  24pi - 18sqrt(3)

Total white area  =  72pi - ( 24pi - 18sqrt(3) )  =  49pi + 18sqrt(3)