All Answers 358087 Answers

+130511

+1

Well.....the circles are tangent.....C lies on the line OP....and C is the point of tangency....so....the image seems to be what you describe.......maybe there's something else to the problem ???

CPhillDec 21, 2019

+118723

0

Hi Madyl,

Looks to me like guest has put in a lot of work.

I don't know why you would ask if the answer came from Mathsisfun. Maybe you have your reasons.

It is good that you have said you do not understand. It is always good to interact with answerers.

Have you spent a lot of time trying to understand?

My criticism is that you have not said thank you (or given a point) for the guests time and attempted explanation.

Do you think guest, or anyone else, should give you more time since you have not shown appreaciation for the time he/she has already spent?

Note:

I have not looked at the question or the answer. I do not know if the answer is correct or if it is well explained.

MelodyDec 21, 2019

+130511

+1

Note that the total possible arrangements are 5! = 120

Let's look at the arrangements where AD do sit together

AD - - -

-AD - -

- - AD -

- - - AD

So they could occupy 4 different positions....and for each of these they could be seated in 2 ways (AD or DA)

And for each of these.....the other 3 can be seated in 3! ways

So.....the number of seatingss where they don't sit together =

120 - (4)(2)(3!) =

120 - 8*6 =

120 - 48 =

72

CPhillDec 21, 2019

#3

+288

+1

Did you copy this from Math is Fun? I can't really understand this solution.

madylDec 20, 2019

+1490

+1

PC = 3

Make sure your question is stated correctly. I do not see anny other answer to that question.

If the answer is PC = 12, than your question is NOT correctly written.

I think it's not the line OP; it's rather a line that goes through O and P.

DraganDec 20, 2019

+130511

+2

Sudsw....from what you describe, see the image :

PC = 3

CPhillDec 20, 2019

+130511

0

*******

CPhillDec 20, 2019

+130511

+2

We have the form

P = ab^t where a = the original population = 27 and t is the number of days

So....we have

36 = 27b^3 divide both sides by 27

(4/3) = b^3 take the 3rd root

(4/3)^(1/3) = b

So we want to find this

P = 27[ *4/3)^(1/3) ] ^30

P = 27 (4/3)^10 = exact value

P ≈ 479 bees after 30 days

CPhillDec 20, 2019

#3

+118723

+2

Thanks Chris, I think you overlooked one of the conditions.

I just did this in ful but it was lost.

Anyway here is my pic

\(Area = \frac{135}{360}\times \pi\times 6^2\\ Area = \frac{27\pi }{2} \; units\; squared \)

MelodyDec 20, 2019

+130511

+1

We can find the radius of the circle, R, by using the Law of Cosines

We have that

(9 sqrt (3))^2 = 2R^2 - 2R^2 ( cos 120°)

243 = 2R^2 - 2R^2 * (- 1/2)

243 = 2R^2 + R^2

243 = 3R^2

Divide both sides by 3

81 = R^2

Then the area of the circle =

pi R^2 =

pi *81 =

81 pi units^2

CPhillDec 20, 2019

+130511

+1

Q1

4log₃ c + 4log₃ d = 4 [ log₃ c + log₃ d ] = 4 log₃ (cd) ] = log₃ (cd)⁴ TRUE

(3/4) [ Ln a + Ln b ] = (3/4) Ln (ab) = Ln(ab)^(3/4) = Ln ( ⁴ √ [a³b³ ] ) FALSE

3 Ln e + 2 Ln f = Ln e³ + Ln f² = Ln (e³ f² ) FALSE

Q2

Note that the points (0, 1) and (3, 6.4) are on the original function

So....the points (1, 0) and (6.4,, 3) are on the one of the graphs shown

This has to be the graph on the bottom left

CPhillDec 20, 2019

0

a=7;b=7;c=7;d=7;e=7;f=7;g=7;p=0;cycle:n=a*1000000+b*100000+c*10000+d*1000+e*100+f*10+g;if(n<10000000 and a==g and b==f and c==e, goto loop, goto next); loop:printn," ",;p=p+1; next:g++;if(g<10, goto cycle, 0);g=7;f++;if(f<10, goto cycle, 0);g=7;f=7;e++;if(e<10, goto cycle,0);g=7;f=7;e=7;d++;if(d<10, goto cycle,0);g=7;f=7;e=7;d=7;c++;if(c<10, goto cycle,0);g=7;f=7;e=7;d=7;c=7;b++;if(b<10, goto cycle,0);g=7;f=7;e=7;d=7;c=7;b=7;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT: 7777777 7778777 7779777 7787877 7788877 7789877 7797977 7798977 7799977 7877787 7878787 7879787 7887887 7888887 7889887 7897987 7898987 7899987 7977797 7978797 7979797 7987897 7988897 7989897 7997997 7998997 7999997 8777778 8778778 8779778 8787878 8788878 8789878 8797978 8798978 8799978 8877788 8878788 8879788 8887888 8888888 8889888 8897988 8898988 8899988 8977798 8978798 8979798 8987898 8988898 8989898 8997998 8998998 8999998 9777779 9778779 9779779 9787879 9788879 9789879 9797979 9798979 9799979 9877789 9878789 9879789 9887889 9888889 9889889 9897989 9898989 9899989 9977799 9978799 9979799 9987899 9988899 9989899 9997999 9998999 9999999 Total = 81

P.S. There may be a "bug" in the code !! If so, please let me know. Thanks.

GuestDec 20, 2019

-2

I've added some more points and an extra circle.

If you draw the two tangents from a point outside to any circle, the tangent lengths will be equal. So if G is the midpoint of AB, then GA = GP = GB and a circle centred on G will go through A, B and P.

So APB = 90 and so cos(ABP) = cos(x) = sin(PAB)

Also ABP = BPD by the parallels.

PB = ABcos(x) and PBcos(x) = 0.5PD => PD = 2AB cos^2(x)

Similarly CP = 2AB sin^2(x) => CD = 2AB(cos^(x) + sin^2(x) ) = 2AB

When a line is tangent to a circle (at P, say) and PQ is a chord of the circle, this chord makes an angle across the other side of the circle that is equal to the angle between the tangent and the chord. I have called this the 5th angle property of a circle (because there are 4 others)

So using the property ABP = BDP and also BAP = ACP. I know that APB = 90 (this is the 3rd angle property of a circle, same thread) so BAP + ABP = 90 => ACP + BDP = 90 as well.

So mark H = midpoint CD => AB = HD = CH => ABHC and ABDH are parallelograms.

By the 5th angle property of a circle ABP = BDP and BAP = ACP

So ACP + BDP = 90

triangles ACH, BHD and HAB are congruent (SAA) so AHB = 90 => H lies on the circle through A, B and G => GH = 0.5AB

GuestDec 20, 2019

+1

A coin is flipped eight times, and the sequence of heads and tails occurring is recorded. How many distinct sequences are possible?

256 i.e. 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 also expressed as 2⁸

_.

GuestDec 20, 2019

+1

There are 13*3 possibilities for the first pair, leaving 12*3 possibilities for the 2nd pair, leaving 11*4 possibilities for the singleton. So: 13*3*12*3*11*4 = 61776 in total.

GuestDec 20, 2019

0

There are 7*7*7 = 343 palindromes that can be made.

GuestDec 20, 2019

+130511

+1

This solution can be easily found graphically : https://www.desmos.com/calculator/8tv2ssgvrv

The area = two triangles each with a base of 4 and a height of 2

So.....the area = 2 [ (1/2) (4) (2) ] = 8 units^2

CPhillDec 20, 2019

#5

0

Yup, this also is incorrect when I enter it into the answer box. Thank you for your time and help though!

GuestDec 20, 2019

0

This seems to be incorrect. Thank you though!

GuestDec 20, 2019

+130511

+2

See the graph here : https://www.desmos.com/calculator/qeidqnvtts

The intersection of this line and the circle will form the diameter of the circle....to find the radius we can complete the square on x in the equation of the circle......so we have....

x^2 - 12x +36 + y^2 = 28 + 36

(x + 6)^2 + y^2 = 64

The center is (6, 0)

The radius = sqrt (64) = 8

So....the area we are looking for is

(1/2) pi (8)^2 =

32 pi units^2

CPhillDec 20, 2019

0

Um... Just wondering, are there numbers missing from this post?

GuestDec 20, 2019

#9

+449

+1

Happy Belated B-day!!

SVS2652Dec 20, 2019