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Um, I don't have a picture, so I'm just going to solve assuming there that the garden is a perfect circle with a perfect circle path going around it.

So the garden is obviously pi(4)^2 = 157.9136704174297379 

However, according to my assumptions, the garden path is going around the garden which just forms a bigger circle. So you can just think that there is one big circle with a radius of 6.

pi(6)^2 = 355.3057584392169103

355.3 feet^2 is what I believe is the answer.

If you can, upload the picture or give us a link so we can confirm.

What you can learn from this is that whenever it asks you the probability of something happening then the probability something else happening, you can find the total probability by multiplying both probabilities.

So if a spinner has 10 sections and 2 are blue and 8 are orange.

You will do 2/10 * 8/10 = 16/100 -> 4/25.

Congratulations to both Melody and Cphill!

Another simple way:

ALL the pluses and minues cancel each other out as follows:

-1 + 2 +3 + 4 - 5 -6 -7 -8 -9 +10 +11 + 12 + 13 + 14 + 15 + 16 - 17 - 18.........etc =64 - which is the 4^3 and the LAST cube in all these additions and subtractions. The process holds true for ALL 99 terms, leaving the value of the LAST cube, which is 100. Therefore:

100^3 =1,000,000 - which is the sum of this sequence.

Another way:

Since a(30) - a(25) + 1 = 6 terms which go up by a factor of 10, then:

a(1331) - a(30) + 1 =1,302 terms. Then we simply divide:

1,302 / 6 =217 6-term periods. And finally:

100 x 10^217 = 10^219 - which is a(1,331)'s term.

(4x+3y)(6x-5y) = 24x²-20xy+18xy-15y² = 24x²-2xy-15y²

\(\text{Both points are on the line so plug them in}\\ a(-5+4i) + b(-5-4i)=38\\ a(7+2i)+b(7-2i) = 38\\ (-5a-5b)+i(4a-4b)=38\\ (7a+7b)+i(2a-2b)=38\)

\(-5(a+b)-14(a+b)=-38\\ -19(a+b)=-38\\ a+b=2\)

\(-5(2) + i(4a-4(2-a))=38\\ i(8a-8)=48\\ i(a-1)=6\\ a-1 = -6i\\ a=1-6i\\ b=2-(1-6i) = 1+6i\)

\((a,b) = (1-6i,1+6i)\)

You have asked these questions before Rudram, it is time for you to do some of your own homework.

The answers to some of these can ealilz be found on the net, at least then you will be practicing your research skills.

Do your own homework !

The dot product of orthogonal vectors   = 0

So

4*3 + -7*k  = 0

12 - 7k  = 0

12 = 7k

k  = 12/7

And

2*3 +  (12/7)* m  = 0

6 + (12/7)m  = 0

(12/7)m  = - 6

m = -6(7/12)   =  - 7/2

Hi Guest,

Most of the high end answers come from very professional people with degrees in Mathematics, Physics or Engineering. 

These people are mostly, but not all, retired.

There are other high end answerers who are currently studying these subjects in University.

Then of course there are a smattering of very capable people who are still at school who provide great answers. 

Then you have the school question askers and the people who come here primarily for socialising. 

Most of the askers are at school or doing online courses. 

There are a few regular question askers that are much older, probably retired. They are studying mathematics (maybe on their own) because they enjoy it and it helps keep their minds active.

That is how I see the demographics here but if anyone sees it differently feel free to say so :)

Nah....probably  < 1/24  probability that you are both wrong......

Longer than forever.........!!!!!!!

THX, Melody   !!!!!

Thanks Chris   

Either that or we are both wrong   

Thanks guest for giving me a hint :)

\(x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}\\ x={4\over{(\sqrt5+1)(\root 4\of5+1)(\root 8\of5+1)(\root {16}\of5+1)}}\times \frac{{(\sqrt5-1)(\root 4\of5-1)(\root 8\of5-1)(\root {16}\of5-1)}}{{(\sqrt5-1)(\root 4\of5-1)(\root 8\of5-1)(\root {16}\of5-1)}}\\ x= \frac{4{(\sqrt5-1)(\root 4\of5-1)(\root 8\of5-1)(\root {16}\of5-1)}}{{(5-1)(\root \of5-1)(\root 4\of5-1)(\root {8}\of5-1)}}\\ x= \frac{{(\sqrt5-1)(\root 4\of5-1)(\root 8\of5-1)(\root {16}\of5-1)}}{{(\root \of5-1)(\root 4\of5-1)(\root {8}\of5-1)}}\times \frac{{(\root \of5+1)(\root 4\of5+1)(\root {8}\of5+1)}}{(\root \of5+1)(\root 4\of5+1)(\root {8}\of5+1)}\\ x= \frac{{(5-1)(\root 2\of5-1)(\root 4\of5-1)(\root {16}\of5-1)}}{{(5-1)(\root 2\of5-1)(\root {4}\of5-1)}}\\ x= \frac{{(\root {16}\of5-1)}}{{1}}\\ x=\root {16}\of5-1\\ x+1=\root {16}\of5\\ (x+1)^{48}=(5^{\frac{1}{16}})^{48}\\ (x+1)^{48}=5^3\\ (x+1)^{48}=125\\ \)

Let  3^x  = m      .....so........(3^x)^2  = (3^2)^x   = m^2  = (9)^x

So we have

m^2 - m + 1    complete the square on m

m^2 - m + 1/4 + 1 - 1/4

(m - 1/2)^2 + 3/4

And this will be a minimum  of 3/4   when  m = 1/2

Impressive, Melody   !!!!!

I also worked it out and I also got 1/24 so you are correct

y  = x^2 + a      y =ax

Set these equal  

x^2 + a  = ax

x^2 -ax + a  = 0

The discriminant   is

a^2 - 4a

To have real solutions.....this must be    ≥   to 0....so

a^2 - 4a    ≥  0

a ( a - 4)   ≥ 0  

This will be true for a on these intervals  :    (-inf, 0 ]  U [ 4, inf)

The hint is that  \((a^2+a+1)^2 \ne a^4+a^2+1\)

\(\frac{a}{a^2+a+1}=\frac{1}{6}\\ (\frac{a}{a^2+a+1})^2=\frac{1}{36}\\ (\frac{a}{a^2+a+1})^2=\frac{1}{36}\\ (\frac{a^2+a+1}{a})^2=36\\ \frac{a^4+2a^3+3a^2+2a+1}{a^2}=36\\ \frac{a^4+a^2+1+2a^3+2a^2+2a}{a^2}=36\\ \frac{(a^4+a^2+1)+2a(a^2+a+1)}{a^2}=36\\ \frac{(a^4+a^2+1)}{a^2}+\frac{2a(a^2+a+1)}{a^2}=36\\ \frac{(a^4+a^2+1)}{a^2}+2*\frac{(a^2+a+1)}{a}=36\\ \frac{(a^4+a^2+1)}{a^2}+12=36\\ \frac{(a^4+a^2+1)}{a^2}=24\\ \frac{a^2}{(a^4+a^2+1)}=\frac{1}{24}\\ \)

P(pen from the first box)  = 4/8   =1/2

P ( crayon from the second box)   = 7/12

So

P(pen from the first box  and crayon from the second)   =  (1/2) (7/12)  =  7/24

What is the sum of all possible values of x such that 2x(x-10)=-50? 

Multiply it on out.                                       2x² – 20x = –50

Add +50 to both sides.  This

puts the equation in standard

quadratic equation form.                           2x² – 20x + 50 = 0

This will factor.                                           (2x – 10)(x – 5) = 0

Set each factor to zero.                              2x – 10 = 0     which means one solution is x = 5  

                                                                     x – 5 = 0       which means the other solution is x = 5

Add the two solutions together and

that sum is the answer to the question.        5 + 5 = 10  

Or would this mean that since both solutions

are 5, then the total would be 5.  I think it depends

on how you view the question.  Maybe I'm wrong.

.

We  have that 

(36-20) take only Mathematics = 16

(27-20)  take only Physics  = 7

20 take both....

So  (60) - ( 16 + 7 + 20) =   

60 -  43   =

17  take neither....