All Answers 358184 Answers

We know that

1 + d =  a           (1)

1 + 9d = ar         (2)

1 + 33d = ar^2    (3)

Subbing (1) into (2) we have that

1 + 9d = (1 + d)r   ⇒    r =  (1 + 9d) / ( 1 + d)      (4)

Sub  (1) and  (4)  into (3)

1 + 33d  =  (1 + d)  (1 + 9d)^2 / (1 + d)^2      simplify

1 + 33d =  (1 + 9d)^2 / (1+ d)

(1 + d) (1 + 33d) = (1 + 9d)^2      simplify

33d^2 + 34d + 1 = 81d^2 + 18d + 1

48d^2 - 16d  = 0

16d ( 3d - 1) = 0

Setting both factors to 0 and solving for d  produces d = 0  (reject) or  d = 1/3  (accept)

And    1 + d = a      .....so.....1 + 1/3 = a  = 4/3

And 1 + 9d = ar   

1 + 9(1/3) = (4/3)r  

1 + 3 =  (4/3)r

4 = (4/3)r

r = 3

So....the 34th term   of the arithmetic series  =   1 + 33d   =   1 + 33(1/3)  = 1 + 11 =  12

Its mainly the fact that I have no idea where I want to go in life and yet my high school is adament about me finding something I'll stick with until I retire, and the rising dread of possibly not being able to do the job I aspire to do. Aside from that, theres personal stuff I find people don't really need to know about, especially on a public forum.

But I think thats everyone's life at some point.

Yes    Kinetic   means 'moving' energy

The ball has KE since it is MOVING.

The ball ALSO has GPE since it is 40 m up in the air (see previous posting)

K.E. = 1/2 mv^2  =  1/2  2 kg (5 m/s)^2 =   25  kg-m^2/sec^2 = 25 j

So that means if the ball rolls it has kinetic energy? 

The ball is sitting....not moving...so it has no KINETIC energy...it has  POTENTIAL energy

GPE =    mgh =  2 kg  9.81 m/sec^2  40 m = 784.8  kg-m^2/sec^2 = 784.8 N-m = 748.8 j

I'm in existential dread but otherwise fine, thanks for asking.

(b^2-4ac)=   (25-4(2)(7) )  =    (25-56) = (-31)

There will be two imaginary roots due to the NEGATIVE discriminant

Each minute 1/15  of the tub is filled  and 1/20 of the tub is drained

So.....the portion of the tub that is fillied  every minute is (1/15 - 1/20)  = [ 20 - 15 ]  / 300 =  5 /300 = 1/60 of the tub

Therefore.....it must take 60 minutes to fill he tub when both are opened

First term =1, Common difference =1/3, Number of terms = 34. Here are the terms:

(1, 1.333333333, 1.666666667, 2, 2.333333333, 2.666666667, 3, 3.333333333, 3.666666667, 4, 4.333333333, 4.666666667, 5, 5.333333333, 5.666666667, 6, 6.333333333, 6.666666667, 7, 7.333333333, 7.666666667, 8, 8.333333333, 8.666666667, 9, 9.333333333, 9.666666667, 10, 10.33333333, 10.66666667, 11, 11.33333333, 11.66666667, 12).

The 34th term = 12

The common ratio of the geometric sequence =3

Nth term=F + D*(N - 1), where F=First term, D=Common difference, N=Number of terms

The 4th term in the geometric sequence =3 * 12 =36

36=1 + 1/3*(N - 1), solve for N

N = 106th term.

So, the 4th term of the geometric sequence, or 36, appears as 106th term in the arithmetic sequence.

x3 − 3x2 − 10x

x (x^2-3x-10)

x(x-5)(x+2)

a little tired as usual

How many ways can yo pick two numbers from the list?    7 C  2 = 21 ways to pick two numbers

   only ONE combination , multiplied = 196  (2 x 98) 

  one out of twenty one ways =    1/21

9  x  -4/3  =    (9 x -4)/3 = -36/3 = -12

Partitions of (7) =15 as follows:
7 = 7
6 + 1 = 7
5 + 2 = 7
5 + 1 + 1 = 7
4 + 3 = 7
4 + 2 + 1 = 7
4 + 1 + 1 + 1 = 7
3 + 3 + 1 = 7
3 + 2 + 2 = 7
3 + 2 + 1 + 1 = 7
3 + 1 + 1 + 1 + 1 = 7
2 + 2 + 2 + 1 = 7
2 + 2 + 1 + 1 + 1 = 7
2 + 1 + 1 + 1 + 1 + 1 = 7
1 + 1 + 1 + 1 + 1 + 1 + 1 = 7

Here is a way without calculus :)

Let x = sin θ, y = cos θ.

\(\displaystyle \max_{x,y\in\{x,y:x^2 +y^2 = 1\}} (x+y) = \max_{\theta \in [0,2\pi]} (\sin \theta + \cos \theta) = \max_{\theta \in [0,2\pi]}\left(\sqrt2 \sin\left(\theta+\dfrac{\pi}{4}\right)\right) = \sqrt2\).

So \(\displaystyle \max_{x,y\in\{x,y:x^2 +y^2 = 1\}} (x+y)^2 = \left(\sqrt2\right)^2 = 2\).

Let r be the common ratio.

The series starts like this: 1, r, r², r³, ...

\(r^n = r^{n+1} + r^{n+2}\\ r^2 + r - 1 = 0\\ r = \dfrac{-1\pm\sqrt{5}}{2}\)

Now we found the possible common ratios. 

Denote \(r_1 = \dfrac{-1-\sqrt5}{2},\;r_2 = \dfrac{\sqrt 5 -1}{2}\)

Denote the sum of the series with common ratio r₁ by S₁.

Denote the sum of the series with common ratio r₂ by S₂.

\(S_1 = \dfrac{1}{1-\dfrac{-1-\sqrt 5}{2}} = \dfrac{3-\sqrt 5}{2}\)

\(S_2 = \dfrac{1}{1-\dfrac{\sqrt5-1}{2}} = \dfrac{3+\sqrt5}{2}\)

\(S_1 - r_1 = \dfrac{3-\sqrt5}{2} - \dfrac{-1-\sqrt5}{2} = 2\)

\(S_2 - r_2 = \dfrac{3+\sqrt5}{2} - \dfrac{\sqrt5 - 1}{2} = 2\)

For both cases, we get S-R = 2.

1.

Let a be the first term, r be the common ratio(which is 7/4).

\(b_1+b_2+\cdots+b_{10} = \dfrac{a(r^{10}-1)}{r-1}=180\\ a\cdot\dfrac{\left(\dfrac{7}{4}\right)^{10} -1}{\dfrac{7}{4}-1} = 180\\ a = \dfrac{47185920}{93808891}\)

So b₁ = 47185920/93808891.

b₂ = b₁ * 7 / 4 = 82575360/93808891.

\(b_2 + b_4 + b_6 + b_8+b_{10} = \dfrac{82575360}{93808891} \cdot \left(\dfrac{\left(\dfrac{7}{4}\right)^{10}-1}{\left(\dfrac{7}{4}\right)^2-1}\right)=\dfrac{1260}{11}\)

Therefore the required answer is 1260/11.