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4a^2 + 20ab +25b^2 + 4ac - 10bc + c^2

The factorization of  this isn't easy to see, Manuel....it is this :

(2a + 5b - c)  ( 2a + 5b - c )

We can can check this by  distibuting each term in the first set of parentheses over the terms in the second set of parentheses

4a^2 + 10ab - 2ac

         + 10ab           + 25b^2 - 5bc

                    - 2ac                 - 5bc   + c^2 

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4a^2  + 20ab - 4ac + 25b^2 - 10bc + c^2

So....the factorization is

( 2a + 5b + c)^2

And  taking both roots of this we have

±√(2a + 5b - c )     

3.

f ⋅ g   =   f(x) ⋅ g(x)

                                                Since  f(x)  =  -5x + 3  , we can replace  f(x)  with  -5x + 3

f ⋅ g   =   (-5x + 3) ⋅ g(x)

                                                Since  g(x)  =  6x - 2  , we can replace  g(x)  with  6x - 2

f ⋅ g   =   (-5x + 3) ⋅ (6x - 2)

Before we simplify, let's get the domain.

There are no denominators or anything, so the domain is all real numbers.

f ⋅ g   =   (-5x)(6x) + (-5x)(-2) + (3)(6x) + (3)(-2)

f ⋅ g   =  -30x²  +  10x  +  18x  +  -6           Combine like terms...

f ⋅ g   =  -30x²  +  28x  -  6

Well...one way is to give good answers....if you do - the mods (and others ) - will give you points

Let the number of candy bar packages be  n .

$18  -  $6.50  -  $1.15n   ≥   $5.75

Now we can drop the dollar signs from every number and simplify.

18  -  6.5  -  1.15n   ≥   5.75

11.5  -  1.15n   ≥   5.75

                                            Subtract  11.5  from both sides of the inequality.

-1.15n  ≥  -5.75

                                            Divide both sides by  -1.15  and flip the inequality sign.

n   ≤  5

The number candy bar packages must be less than or equal to  5 . The maximum number is  5  .

Thank you so much!

Looking at triangle  PQR....

RP  =  PQ    so    ∠QRP  =  ∠PQR  =  x°

x + x + 3y   =   180

2x + 3y   =   180

Looking at triangle SPQ....

∠SPQ  =  180 - 3y

x + 2y + (180 - 3y)  =  180

x - y + 180   =   180

x - y   =   0

x   =   y

Substitute  y  for  x  in the equation  2x + 3y  =  180

2y + 3y  =  180

5y  =  180

y  =  36

3y  =  3 * 36

3y  =  108

I got it it is 15 :D

I actually managed to figure this one out!  Thanks so much though!

show me a EXPANDED FORM YEAL!

24 is the answer

thank you very much Alan for clearing this to me 

Solution: 

\(\text {With six dice }\\ \small \text {There are } \dbinom{6}{2} \text { ways to chooses two dice for the “ones” }\\ \small \text{There are then} \dbinom{4}{2} \text { ways to choose two dice for the “twos” }\\ \small \text{There are } 6^2 = 36 \text{ ways to roll two dice. There is one way to roll two dice for two 1s (snake eyes) (1/36). }\\ \small \text{There is one way to roll two dice for two 2s (1/36).} \\ \small \text{Of the remaining two dice, there are 4 of 6 ways to roll each die without a 1 or 2 (2/3). }\\ \dbinom{6}{2} * (1/36) * \dbinom{4}{2} * (1/36) * (2/3) * (2/3) = 5/162 \\ \)

We really do get it right if we monkey around with it enough.

GA

1. On the label of a bottle of vinegar, the information appears: "acetic acid solution at 4% by weight". If the bottle contains 750mL of vinegar and its density is 1.006g / mL, calculate the mass of acetic acid contained in the bottle.

It is easy, you can cancel the mL units out ! 

The only other thing you need to know is that mass can be measured in grams

\(Total\;mass\;=750mL*\frac{1.006g}{1mL}=750*1.006g = 754.50g\\ Mass\;of\;acid\;\;=\frac{4}{100}*754.5g=30.18grams \)

2. A hospital organized a blood donation event where 80 pints of blood were collected. The hospital expected to get 8 gallons of blood with the event. Has the goal been met?

Why would you convert to metric?

There are 8 pints in a gallon   (I expect you are from the U.S. and you are supposed to know this)

so 8 gallons = 64 pints

They expected 64 they got 80 so YES they were happy :)  The goal was not only met it was exceeded. 

Please note that I have edited my answer and it is open to examination  

.......

Here's my reasoning......

1st Flip....possible scenarios

( Starting Floor   ⇒ Ending Floor )

6 ⇒ 7

6 ⇒ 5

Second Flip ....possible scenarios

7 ⇒ 8

7 ⇒ 6

5 ⇒ 6

5 ⇒ 4

Third flip - possible scenarios

8 ⇒ 9        

8 ⇒ 7        

6 ⇒ 7     [ this one is no good...being at the 6th floor intially means that she was below the 7th floor on one of the flips....the requirement is being at 7 or higher on all 5 flips ]

6 ⇒ 5

4 ⇒ 5

4 ⇒ 3

Fourth Flip - possible scenarios

9 ⇒ 10         5 ⇒ 6

9 ⇒ 8           5 ⇒ 4

7 ⇒ 8           3 ⇒ 4

7 ⇒ 6           3 ⇒ 2

Last Flip - possible scenarios

10 ⇒ 11       6 ⇒ 7       2 ⇒ 3

10 ⇒ 9         6 ⇒ 5       2 ⇒ 1

8 ⇒ 9           4 ⇒ 5

8 ⇒ 7           4 ⇒ 3

Total possible scenarios  =   2 + 4 + 6 + 8 + 10   =  30

Notice, however, that only 11 of these guarantee a floor of 7th or higher on each of  5 flips

Favorable flip 1 outcomes

6 ⇒ 7

Favorable  flip 2 outcomes

7 ⇒ 8

Favorable Flip 3 outcomes

8 ⇒ 9

8 ⇒ 7

Favorable Flip 4 outcomes

9 ⇒10

9 ⇒ 8

7 ⇒ 8

Favorable Flip 5 outcomes

10 ⇒ 11

10 ⇒ 9

8 ⇒ 9

8 ⇒ 7

So.......the probability  that she is on a floor 7th or higher on each of the 5 flips  is

11 / 30

Three concentric circles are drawn such that the area of the smallest circle is equal to the area of each of the rings.
If the radius of the largest circle is 12 cm,
what is the radius of the smallest circle?

\(\begin{array}{rcll} \text{Let smallest circle radius $=r_s$ } \\ \text{Let middle circle radius $=r_m$ }\\ \text{Let largest circle radius $=r_l$ }\\ \end{array} \)

\(\begin{array}{|lrcll|} \hline & \pi r_s^2 = \pi r_m^2-\pi r_s^2 = \pi r_l^2 - \pi r_m^2 \quad & | \quad : \pi \\ & r_s^2 = r_m^2- r_s^2 = r_l^2- r_m^2 \\ \hline \end{array} \)

\(\begin{array}{lrcll} (1) & 2r_s^2 &=& r_m^2 \\ \hline (2) & 2r_m^2 &=& r_l^2 + r_s^2 \quad & | \quad r_m^2 = 2r_s^2 \\ & 2\cdot (2r_s^2) &=& r_l^2 + r_s^2 \\ & 4r_s^2 &=& r_l^2 + r_s^2 \\ & 3r_s^2 &=& r_l^2 \quad & | \quad \sqrt{ }\\ & \sqrt{3}r_s &=& r_l \quad & | \quad r_l = 12 \\ & \sqrt{3}r_s &=& 12 \\ & r_s &=& \dfrac{12}{\sqrt{3}} \\ & r_s &=& \dfrac{12 \sqrt{3}}{\sqrt{3}\sqrt{3}} \\ & r_s &=& \dfrac{12 \sqrt{3}}{3} \\ & \mathbf{r_s} &\mathbf{=}& \mathbf{4 \sqrt{3} } \\ \end{array}\)

\(\text{The radius of the smallest circle is $\mathbf{4 \sqrt{3} }$ cm $\approx \mathbf{6.9} $ cm } \)

A.

60 cm  -  2 cm/min *  5 min   =   60 cm - 10 cm   =   50 cm

8 cm  +  1.25 cm/min * 5 min   =   8 cm  +  6.25 cm   =   14.25 cm

B.

water level of old aquarium  =  W  =  60 - 2T

water level of new aquarium  =  W  =  8 + 1.25T

C.

water level of new aquarium  >  water level of old aquarium

8 + 1.25T  >  60 - 2T

D.

8 + 1.25T  >  60 - 2T

3.25T  > 52

T > 16

E.

The new aquarium's water level will be greater than the old aquarium's water level after  16  mins.

F.

Find the time it takes for the water level of the old aquarium to be  0  .

0  =  60  - 2T

2T  =  60

T  =  30

Find the water level of the new aquarium at this time.

W  =  8 + 1.25(30)

W  =  45.5

Mmmm....that problem seems familiar!!!!  [inside joke ]

Good job, hectictar  !!!

.........

*edit again* I just thought of this....

area of small circle  +  area of middle ring + area of outer ring   =   area of big circle

area of small circle + area of small circle + area of outer ring   =   area of big circle

area of small circle + area of small circle + area of small circle   =   area of big circle

3(area of small circle)   =   area of big circle

3π r²  =  π * 12²      ,  where  r  is the radius of the small circle.

3π r²  =  144π

                               Divide both sides by  3π .

r²  =  48

                               Take the positive square root of both sides.

r  =  4√3

correction and the sum: