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Uhh... if you take 2 and there are 2 on the counter.

You have 0 left?

Two, since you take two.

Lol

Answer: 42.05 (units)

Explanation:

If AE is perpendicular to BC and CE = 20 with AE = 21, then AC = 29 by Pythagorean Theorem. 

Now, call AB x and BE y.

\(21^2+y^2=x^2\)

\(29^2+x^2=(y+20)^2\)

by Pythag.

Simplifying things...

\(x^2-y^2=441\)

\(x^2-y^2=40y-441\)

So \(441=40y-441, y=22.05\).

To find the lengh of BC, it's just BE+EC, or 22.05+20 = 42.05 un.

You are very welcome!

:P

Agreed!

We have

(2x^2  - 3y)^3  =

C(3,0)(2x^3)^3 * (3y)^0   -  C(3,1) (2x^3)^2(3y)^1  + C(3,2)(2x^3)^1 (3y)^2  + C(3,3)(2x^3)^0 (3y)^3  =

   1   (8x^9)   -   3 (4x^6)(3y)  +  3 (2x^3)(9y^2)  + 1 ( 27y^3) 

The coefficients are in  red

And the final expansion is

8x^9  - 36x^6y + 54x^3y^2 - 27y^3

Very nice, heureka  !!!

The last answer is correct.....the series sum is (3/4)

x = speed of slower car

x+10 = faster car

rate  *  time = distance

(x   + x+10 )  * 2 hrs = 280 miles    solve for 'x' = slower car   then   x+10 = faster car

Let the width of the path  = W

So we have that

(18 + 2W) ( 22 + 2W)  =  896      simplfy

396  + 44W + 36W  + 4W^2  =  896

4W^2  + 80W + 396  = 896     subtract 896 from both sides

4W^2  + 80W  - 500  = 0       divide through by 4

W^2  + 20W  -  125   = 0       factor as

(W - 5) ( W + 25)  = 0

Setting each  factor to 0  and solving for W produces  W = 5    or W  = -25

Obviously....only the positive answer makes sense....so...

The width of the path  = 5 ft

SA = πr² + πrl

 60 pi   =   pi r^2 + pi r l

  60      =      r^2   + r l

  60      =      4^2   + 4 l

   44 = 4 l

    l = 11 cm

Physiological dormancy prevents embryo growth and seed germination until chemical changes occur. Physiological dormancy is indicated when an increase in germination rate occurs after an application of gibberellic acid (GA3) or after Dry after-ripening or dry storage.

There are 450 seats in the lower level of a concert hall with b balcony seats in the upper level. So far, 170 tickets have been sold, which is 1/5 if the total number of seats in the concert hall. How many balcony seats are in the concert hall?

1/5 x = 170     solve for x   the total seats in the concert hall     then    x - 450 = balcony seats.

Annette needs 20 gallons of water with a 3.5% salt concentration for her fish tank. She has a 2% salt solution and a 10% salt solution. How many gallons of each solution does she need for her tank?

If the car decreases at the rate of 20% per year, it retains 80% of its value.

So, after 6 years, it will be worth 20000 x .80⁶  =  5242.88.

I leave the explanation to you.

If each child can have 1 or more chocolates, then you have:

[6 - 1] C [3 - 1] =5C2 =10 ways.

a=2;b=2001;d=0;f=1;cycle:d=((d+a)/2);a++;if(a<=b, goto cycle,0);printd+2002

OUTPUT = 4,002

x = 3.5% solution amount

20-x = 2% amount

salt content of ingredients     = salt content of final solution

3.5% (x)   + 2% (20-x)   =  10% (20)

.035 x      + .02 (20-x)    = .1 (20)                       solve for "x"  the amount of 3.5%       then  the amount of 2%  =  20-x   gallons

 If they use ALL of the ingredients:

2 * $ 3.72    +  4 * $ 4.33  +  4* $4.15

____________________________            =      .........

(2              +   4             +     4)    

89 = 3.5n + 8   |-8

81 = 3.5n         | :3.5

  n = 6

The back of Dante's property is a creek. Dante would like to enclose a rectangular area, using the creek as one side and fencing for the other three sides, to create a corral. If there is 760 feet of fencing available, what is the maximum possible area of the corral?

Hello sandwich!

\(A= a\cdot b\\ a+2b=760\ ft\\ 2b=760\ ft-a\\ b=380\ ft-\frac{a}{2}\)

\(A=f(a)=a\cdot (380\ ft-\frac{a}{2})\\ A=f(a)=-\frac{a^2}{2}+380\ ft\cdot a\\ A'= \frac{df(a)}{da}=-a+380\ ft=0\)

\( \color{blue}a=380\ ft\\ b=380\ ft-\frac{a}{2}=380\ ft-190\ ft\\ \color{blue}b=190\ ft\)

Dante creates a rectangular field 380ft * 190ft.

  !

Find the sum \(\dfrac{2^1}{4^1 - 1} + \dfrac{2^2}{4^2 - 1} + \dfrac{2^4}{4^4 - 1} + \dfrac{2^8}{4^8 - 1} + \dots\)

\(\begin{array}{|rcll|} \hline && \dfrac{2^1}{4^1 - 1} + \dfrac{2^2}{4^2 - 1} + \dfrac{2^4}{4^4 - 1} + \dfrac{2^8}{4^8 - 1} + \dots \\\\ &=& \dfrac{2^1}{2^2 - 1} + \dfrac{2^2}{2^4 - 1} + \dfrac{2^4}{2^8 - 1} + \dfrac{2^8}{2^{16} - 1} + \dots \\\\ && \boxed{ \text{We substitute }y =2 } \\ &=& \dfrac{y^1}{y^2 - 1} + \dfrac{y^2}{y^4 - 1} + \dfrac{y^4}{y^8 - 1} + \dfrac{y^8}{y^{16} - 1} + \dots \\\\ && \boxed{\dfrac{y^1}{y^2 - 1}\\ =\dfrac{1+y-1}{y^2-1}\\ = \dfrac{1+y}{y^2-1} -\dfrac{1}{y^2-1}\\ = \dfrac{1+y}{(y-1)(y+1)} -\dfrac{1}{y^2-1} \\ = \dfrac{1}{y-1} -\dfrac{1}{y^2-1}\\ \mathbf{\dfrac{y^1}{y^2 - 1}} = \mathbf{\dfrac{1}{y-1} -\dfrac{1}{y^2-1}}} \\ &=& \left(\dfrac{1}{y-1} -\dfrac{1}{y^2-1}\right) \\ & +& \left(\dfrac{1}{y^2-1} -\dfrac{1}{y^4-1}\right)\\ & +& \left(\dfrac{1}{y^4-1} -\dfrac{1}{y^8-1}\right) \\ &+& \left(\dfrac{1}{y^8-1} -\dfrac{1}{y^{16}-1}\right) + \dots \quad | \quad \text{The sum is a telescoping one} \\\\ &=& \dfrac{1}{y-1} \quad | \quad y=2 \\ &=& \dfrac{1}{2-1} \\ &=& \dfrac{1}{1} \\ &=& \mathbf{1} \\ \hline \end{array} \)

\(\dfrac{2^1}{4^1 - 1} + \dfrac{2^2}{4^2 - 1} + \dfrac{2^4}{4^4 - 1} + \dfrac{2^8}{4^8 - 1} + \dots = \mathbf{1}\)

Find  \(f(k)\) such that \(\sum \limits_{k=1}^n f(k) = n^3\).

Not typos, the latex has just been omitted and guest is too lazy to even notice.

Thanks Heureka :)