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how old are you?

Part A.....let P be the number of people that will show up.....so....

The total amount of broccoli needed (in ounces)  =    8P  ounces

Part B

32  =  8P       divide both sides by 8

4 = P            so.....4 people can be fed.....!!!

A test i'm practicing

Part A.....let x be the amount he earned on Sunday.....so

Amount  earned on Saturday  +  amount earned Sunday   =

11  +  x        =    

Represents the total amount he earned

Part B

Let N be the number of dogs that Kyle walked

Then,  Larry walked  2N + 4   dogs

i do not know

161 and a half 

 same here

you are great at math CPhill

weiner=4

was that question for school?

Cool!

if x = y, y = zm, mz = uii, u = i, i = pi, what is x

uii   = pi^3   =   mz   =  y  = x

Yay thanks!! I understand it more too now that you showed how! :D

Oh okay.....

Volume  =  (4/3) * pi * (diameter/2)^3 .......so we have

(4/3) * pi * (11/2)^3  ≈  696.9 cm^3

K why?

Look at the following diagram, NSS......

Triangle DPC  similar to triangle APB......therefore

DP/PC  = AP/PB   .... so.....

DP*PB  = PA*PC

So.....the theorem is that when two secants are drawn from an external point outside a circle, the product of the length of one secant and its external segment  equals the product of the length of the other secant and its external segment

Applying this to your problem we have that

AC*CB = (CD + x) * CD           Note......AC  = AB + BC

(AB + BC) * BC  = (CD + x) * CD

(34 + 12) * 12  =  (8 + x) * 8        simplify

(46)*12  =  64 + 8x

552 = 64 + 8x       subtract 64 from both sides

488 = 8x               divide both sides by 8

61  = x

[Looks as though you got the correct answer !!! ]

Max: Check your answer against this step-by-step answer here:

http://www.integral-calculator.com/

78/483 = 0.1614906832298137

Take the integral:
 integral x^3 sqrt(9 - x^2) dx

For the integrand x^3 sqrt(9 - x^2), substitute u = x^2 and du = 2 x dx:
 = 1/2 integral sqrt(9 - u) u du
For the integrand sqrt(9 - u) u, substitute s = 9 - u and ds = - du:

= 1/2 integral(s - 9) sqrt(s) ds
Expanding the integrand (s - 9) sqrt(s) gives s^(3/2) - 9 sqrt(s):
 = 1/2 integral(s^(3/2) - 9 sqrt(s)) ds
Integrate the sum term by term and factor out constants:
 = 1/2 integral s^(3/2) ds - 9/2 integral sqrt(s) ds
The integral of s^(3/2) is (2 s^(5/2))/5:
 = s^(5/2)/5 - 9/2 integral sqrt(s) ds
The integral of sqrt(s) is (2 s^(3/2))/3:
 = s^(5/2)/5 - 3 s^(3/2) + constant
Substitute back for s = 9 - u:
 = 1/5 (9 - u)^(5/2) - 3 (9 - u)^(3/2) + constant
Substitute back for u = x^2:
 = 1/5 (9 - x^2)^(5/2) - 3 (9 - x^2)^(3/2) + constant
Which is equal to:
Answer: | = -1/5 (9 - x^2)^(3/2) (x^2 + 6) + constant

13824 

whatnot

24^3 = 13824

9z-10z=-6-1

-z=-7

z=7