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im actully thinking the second question is actually C , r=0.8 . the series dverges

Assuming the survey is truly representative of the totals

sample 1   shows    7 out of 12  omelets were H&C    7/12 of 200 = 116.67  on day 1

sample2                  11 out of 18                                11/18 of 200 = 122.22

Average   =  (116.67+122.22)/2 =~~120 per day are H&C

this one is way more simple than i thought! Thank you for your help on this one. (:

1   every number has equal chance      there are 10000 numbers     so   1/10000 chance =.0001

2   there are 24 arrangements of the numbers   3 0 7 9     24/10000        .0024

actually i think the second question is D converges r=1/7

First one

The 4th term in the expansion  of     [ (2x^2y^3)  +  y ]^7   is given by

C(7, 3) * (2x^2y^3)^(7 - 3) * ( y)^3  =

C(7,3) * ( 2x^2y^3)^4 * y^3   =

35 * 2^4  *  (x^2 y^3)^4 * y^3  =

35 * 16  *  x^8 * y^12 * y^3  =

560 x⁸ y¹⁵

Here's the complete expansion :

128 x^14 y^21 + 448 x^12 y^19 + 672 x^10 y^17 + 560 x^8 y^15 + 280 x^6 y^13 + 84 x^4 y^11 + 14 x^2 y^9 + y^7

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so sorry I forgot to reply, but yes this makes sense! I will add notes with this,thank you so much.

Two good reasons to not be out there kiting checks !  Yikes   

That's how you write in physics.

49.13   - 2(32.50)  + 2(74.25)    = 

49.13  -  65  + 148.50  = 

$132.63 ending balance

d = 3q

n = 2 * 3q -4

q

25 q  + 10 * 3q  + 5 (2*3q-4) = 2020

25q + 30 q   + 30q -20 = 2020

85q = 2040

q = 24

then d = 3q = 72  dimes =  $7.20

See the following image

Note that OXQY  forms a quadrilateral

Since  QX and QY   are tangent to the circle, then angles  QXO  and QYO   = 90 °

And by symmetry,  angle   XQY  = 2(32)    = 64°

And the interior angles of a quadrilateral sum to 360°

So

QXO + QYO + XQY  + XOY  =  360

  90  +   90   +   64  + XOY  =  360

244 + XQY  = 360         subtract  244 from both sides

XOY  = 116°

And since  XOY  is a central angle........then minor arc  XY has the same measure  =   116°

thanks

Tan 52 = opp/adj = opp/62

1.2799 (62) = ~ 79.4 ft tall

195 + 10  ≤  40n  

The ostrich gets a 44f/s x 60s =  2640 ft head start

the cheetah is running   110 - 44   66 f/s faster

    to cover the 2640 ft   the cheetah will need       2640 ft/66f/s = 40 seconds to catch the ostrich

In 60 seconds, the ostrich will run  60*44   = 2640 ft

And let  S  be the number of seconds that it will take the cheetah to catch the ostrich

So....the distance that  the ostrich will cover before being caught  is  2640 + 44S

So....we need to solve this

110S  = 2640 + 44S        subtract  44S from both sides

66S  =  2640      divide both sides by 66

S = 40 seconds     

I'm sorry, could you include a solution? I'd like to understand how to solve the problem on my own.

yes     key in the number of which you want the square root

               then hit the  \(\sqrt{x}\)    key     then the ' = '  key

2 +3 +4 +5 +6 +7 +8

The shorrtest distance is 10 - 5*sqrt(2).

The three angles must sum to 180 in a triangle

6 parts + 3 parts + 7 parts = 16 parts

6/16  x 180 =middle angle

3/16  x 180 = smallest angle

7/16  x 180 = largest angle

The unknown, x, is how many papers he delivered

    money from papers = .10 * x = ,1x

$ 5 + .1x  = 20.60     Solve for x to find the number of papers he delivered  (the Q does not ask for this)

THANK YOU!!!