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5*(-6)+5*(-5)

-30   +  -25= -55

(probably wrong, never confident in the math dept. lol)

Thank you maybe i'll eventually understand this.. >.<

5)  448n^4 - 56n^3 -56n^2 + 7n

Let's factor the last two terms, first  = -7n ( 8n - 1)

Now ,RP..let's guess that  (8n - 1)  might factor the first two terms in some manner

So  we have  (8n - 1)    and notice that if I multiply this  by 56n^3 ...   we have   56n^3 (8n -1)  = 448n^4 - 56n^3

So...the factorization becomes

56n^3 (8n - 1)  - 7n (8n - 1)  =

(8n- 1) (56n^3 - 7n)     and we can factor  a 7n  out of the second factor giving us

(8n - 1) * 7n * ( 8n^2 - 1)  =

7n (8n - 1) (8n^2 - 1)

6) 45a^5 + 54a^4 - 60a^3  - 72a^2

GCF  45a^5  and 54a^4  =  9a^4

GCF -60a^3 and - 72a^2  = -12a^2

So we have

9a^4 ( 5a + 6) - 12a^2 (5a + 6)

(5a + 6) (9a^4 - 12a^2)....we can factor  3a^2  out of the second factor

(5a + 6) * 3a^2 * ( 3a^2 - 4)  =

3a^2 (5a + 6) (3a^2  - 4)

Thank youuu

First one :

We have a right triangle  as a base...the two legs are 6 and  8....the area of a right triangle  = (1/2) (product of the legs) =

(1/2) (6 * 8)  = (1/2) (48)  = 24 m^2  = base area

And the height of this triangular pyramid is 8m

So...the volume is  (1/3) (base area ) ( height) =  (1/3) (24) (8)  = (8) (8)  =  64 m^3

Second one :

If you look at the front face of this prism, it's just a trapezoid with a height of 4.3 cm and bases of 9cm  and 4 cm

The area of this trapezoid  is   (1/2) * height * (sum of the bases)  = (1/2)(4.3) ( 9 + 4)  =  (1/2) (4.3) (13)  = 27.95 cm^2

Now...if we consider all these similar trapezoids to  be  "stacked" to a depth of  6cm...the volume becomes

6 (27.95 )  =  167.7 cm^3

Ruan began painting a room at 9:00 AM. Leonardo, who can paint twice as fast as Ruan, started helping Ryan at 9:20 AM, and they worked together until the room was fully painted at 10:00am. What fraction of the room had been painted by 9:30 AM? Express your answer as a common fraction in lowest terms and clearly show/explain your reasoning.

Let the fraction of the room that Ruan can paint in one minute be  (1/x)

Let the faction of the room that Leonardo can paint in one minute  be 2 (1/x) =  (2/x)

So...fraction of the room painted in one minute * minutes on the job  =  part of the job done

Ruan paints for 60 minutes when the job is completed and  Leonardo paints for 40 minutes when the job is done....so we have this equation :

(1/x)(60) + (2/x)(40)  = 1

60/x  + 80/x  = 1

[60 + 80] / x   = 1

140 / x  = 1    ⇒   x  =  140

So....Ruan can paint (1/140) of the room in one minute 

So...in 30 minutes, Ruan paints  (1/140)(30)  =  30 / 140 of the room

And in  10 minutes, Leonardo paints  (2 / 140)(10) = 20 / 140 of the room

So...together....they paint   (30 + 20 ) / 140   =  50 / 140    = 5 / 14  of the room by 9:30 AM

Formula for sphere volume  = (4/3) pi *  (diameter / 2)^3...so we have....

Volume of outer sphere  =  (4/3) pi * ( 2.8 / 2)^3  

Volume of inner sphere  = (4/3) pi * ( 2.1 / 2)^3

So...area between the spheres is   (4/3) pi  [  (2.8 / 2)^3  -  (2.1 / 2)^3 ] ≈ 6.64 m^3

3)  36x^3  + 96x^2  + 42x + 112

Take out the GCF  between  36x^3  and 96x^2  =  12x^2

Take out the GCF between 42x  and 112   = 14

So we have

12x^2 ( 3x + 8)  + 14 (3x + 8)  .....  (3x + 8)  is common...so....

(3x + 8) (12x^2 + 14)     ....note that we can factor a 2 out of the second factor...and we have...

(3x + 8)* 2* (6x^2 + 7)  =

2(3x + 8) (6x^2 + 7)

4)  42n^3 - 48n^2  - 336n + 384

Take out the GCF between 42n^3 and 48n^2  = 6n^2

Take out the GCF between 336n  and 384  = 48

So we have

6n^2 ( 7n - 8) - 48(7n -8)

(7n -8) (6n^2 - 48)     ...and we can factor a 6 out of the second factor

(7n - 8) * 6 * ( n^2 - 8)

6(7n - 8) (n^2 - 8)

I get 5/14 which is the same as our guest.

Ruan began painting a room at 9:00 AM. Leonardo, who can paint twice as fast as Ruan, started helping Ryan at 9:20 AM, and they worked together until the room was fully painted at 10:00am. What fraction of the room had been painted by 9:30 AM? Express your answer as a common fraction in lowest terms and clearly show/explain your reasoning.

---------------------------------

Here is my logic   

Let Ruan paint one unit of area every minute.

then Leonado is twice as fast so he will paint 2 units of area ever minute.

In the whole hour Ruan will paint  60 units of area

Leonado only paints for 40 minutes so he will paint a total of 80 units of area.

So in the hour they paint 60+80 = 140 units of area and this is the whole room.

Now how much is painted by 9:30am

Ruan has painted for 30 minutes so he has painted  30 units of area.

Leonado has painted for 10 minutes so he has painted 20 units of area.

So in the first half hour  30+20 = 50 units of area is painted.

So that is    \(\frac{50}{140}=\frac{5}{14}\)       of the room is painted by 9:30

Thank you so much!

Thank you so much for your help, but I still don't understand this >.<

help wiith these

"both came out wrong" is neither nice nor mean. It is certainly not polite though.

Plus, I had not idea what you were talking about. At least you have now explained better why you think it was wrong.

Maybe it is your program that has the wrong answer, did you ever consider that?

I'll take (a) look at a for you.

P = 9+9+13+13+7+7+   the arc

= 58mm + the arc

The arc is from a circle with radius9 and subtended from an angle of 44 degrees

so

 \(length \;\;of \;\;arc\;=\frac{44}{360}*2*\pi*9\\ length \;\;of \;\;arc\;=\frac{22}{10}*\pi\\ length \;\;of \;\;arc\;=2.2\pi \approx 6.9mm \)

\(Perimeter=(58+2.2\pi)mm \approx 64.9mm\)

I think CPhill misinterpreted what a composite figure is but he did all the hard bits, you should have been able to figure out why his answer was not correct yourself.  Did you even try to?

When I put the answer in it came out wrong. I do appreciate CPhill for answering my questions. I was telling him that it came out wrong; in a nice way not mean. 

Obisiously my every day is too good..I really appreciate you although good to know that you attend your classed, American studies is one of the my favorite subject.

Hey, I'm Rock and I'm too good..Now tell me about you, I go through your different post in this community ,that's all are quite tremendous.

V=pi x r^2 x h
V=pi x 8^2 x 3
V=192 in^3 before increase.
Let the increase =n
Pi x (8+n)^2 x 3=Pi x 8^2 x (3+n), solve for n
3 π (n + 8)^2 = 64 π (n + 3)

Divide both sides by π:
3 (n + 8)^2 = 64 (n + 3)

Expand out terms of the right hand side:
3 (n + 8)^2 = 64 n + 192

Subtract 64 n + 192 from both sides:
-192 - 64 n + 3 (n + 8)^2 = 0

Expand out terms of the left hand side:
3 n^2 - 16 n = 0

Factor n from the left hand side:
n (3 n - 16) = 0

Split into two equations:
n = 0 or 3 n - 16 = 0

Add 16 to both sides:
n = 0 or 3 n = 16

Divide both sides by 3:

n = 16/3 =5 1/3 inches

ok, this one is a bit involved.  6 Mars like planets that use 1 colony unit, 5 Earth like planets that use 2 colony units

12 total colony units.

First off we note that we only consider combinations with even numbers of Marslike planets.  Otherwise we would have a colonly unit left over.  We also note that we must colonize at least 2 Martian like planets and can only colonize up to the 6 existing Marslike planets.  Below is a table summarizing the results.

Totalling the right column we see there are 100 different combinations using all 12 of the colony units.

I think there is a minor typo in Rom's answer of (30C3 x 20C2) =771,400.

There are \(\dbinom{30}{3}\dbinom{20}{2} = 771400\) different combinations without regard to order

I think it's easiest to count the ways that they can consecutively sit together and subtract this from the total number of seating arrangements.

There are 6 groups of 3 consecutive seats.  Within each group there are 3!=6 ways the three fussy people can be arranged.

Then the remaining 5 people can be arranged 5!=120 ways in the remaining seats.

So there are 6*6*120 = 4320 arrangements where they DO sit consecutively

There are 8! = 40320 total arrangements.

So there are 40320 - 4320 = 36000 arrangements where the 3 do not sit consecutively.

|x+1| = x - 1   has no solution.

without being able to distinguish individual frogs we only have 2 possible orderings

ggbrrr, and rrrbgg

Now if you say we are able to uniquely identify each frog then we have 2!=2 ways of arranging the green frogs and 3!=6 ways of arranging the red frogs.  So there are a total of 2*2*6 = 24 ways of arranging the frogs to their liking.