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Oh, I see. 

Broth         Carrots            Beans

750            1/2                  300

600            2/5                  240

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The above is the graph of \(y=sin(x)-2cos(x)-tan(x)\)over the interval (0, 2pi). There are two zeros: 3.85300 rads and 5.21133 rads. Neither zeros are the size you could fit in a right triangle. So there is something lacking or false in the statement of this problem. Please correct it before others waste time on it. Thanks.

Show work instead of guessing. We get nothing when you guess

8 7/12 of the board is left.

you would have baked 6 cakes. :)

Show your work

Jordy bought 101 stamps.

x equals 4

We have to crate a ratio table

30*50 = 1500

15 dollars is the cost of the box, you have to sell it for 60 dollars to make a 45 dollar profit. 

60/50 = 1.25

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x^2 + y^2 = x + y + 4

x^2 - x + y^2 - y = 4

(x-.5)^2 + (y-.5)^2 - .5 = 4

(x-.5)^2 + (y-.5)^2 = 4.5

The largest value of x is sqrt(4.5) + .5

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The dudemanguyperson has already answered. 

You should respond to him/her.

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I'm not sure what you mean by creating a table.

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Can you also create a table?

600/750 = 4/5

The number of beans is 4/5*300. 

The number of carrots is 1/2*4/5. 

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The perpendicular slope of 2 is -1/2, and the x intercept is 2. 

y= -1/2x + 2

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sqrt((2x + 3)/4)=2sqrt((2x + 3)/4)

Solve for x. :))

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1/2 change of tail. 

1/6 chance of 6. 

1/2*1/6 = 1/12

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That is wrong  :(

Let H, D, and A be the events that the two cards selected are both hearts, diamonds, or aces, respectively. First note that these events are disjoint; the two cards cannot simultaneously be two aces and two diamonds or two hearts (there is only one ace of diamonds and only one ace of hearts). So

                                     \(p(H\ or\ D\ or\ A)= p(H)+p(D) + p(A)\)

Now

                                       \(p(H)=p(D)=\frac{C(13, 2)}{C(52, 2)}=\frac{1}{17}\), and

                                       \(p(A)=\frac{C(4, 2)}{C(52, 2)}=\frac{1}{221}\). So,

                                       \(p(H\ or\ D\ or\ A)=\frac{1}{17}+\frac{1}{17}+\frac{1}{221}=\frac{27}{221}.\)

36  ft *  .305    =   10.98  m  =   radius  of  first pool in meters

22.14  / 2     =    11.07  m  =   radius  of second pool

No  computations needed......the second pool  will  have  the greater  circumference

Total  volume

Bottom  rectangular prism  +  Top rectangular prism

(21 * 7 * 2)   +  (10 *12 * 7)  =  1134  in^3

We have  the  followig 4  possible  systems   and solutions

1 - 3x  +  d  =  3x + 9

3x + 9 +  d  =  4x  - 12

x = -29/5     d  =  -134/5       (  reject.....d  added  to   abs ( 1 - 3x )   is  negative....but  the  second term, abs (3x + 9), is positive  )

3x  - 1  +  d  =  3x + 9

3x + 9  +  d   =  4x - 12

x = 31     d  =  10

1 - 3x  + d   =  3x + 9

3x + 9  + d  =  12  - 4x

x = -5/13    d  =  74/13

3x - 1  + d   =  3x + 9

3x + 9 + d  =  12 - 4x

x =  -1     d   =  10      (reject....iif  x  = -1, the  terms  would  be  4, 6 , 16....not  arithmetic )

The  least possible  value for  x  =  -5/13

Here is visual confirmation that R = 15⁵/₈ is correct.

The answer is yes; line segments AD, BE, and CF would have to be medians, which would by definition make X the centeroid, if the following pairs of triangles have the same area

:\(\Delta AXF\ and\ \Delta FXB\)

\(\Delta AXE\ and\ \Delta EXC\)

\(\Delta BXD\ and\ \Delta DXC\)

Let's look at the reasoning for the last pair; the other two would be similar.

The line segment \(X{H}_{3}\)is perpendicular to BC, so it must be perpendicular to BD and DC both; So it would serve as a heigt for triangles BXD and DXC. If the two triangles have the same height and the same area, it follows that they have the same base; that is, the line segments BD and DC must have the same length. In other words AD is a median.