All Answers 358087 Answers

Looks like you know HOW to solve it, you just made a small mistake....

   See that '1' on the right side of the equation?  It needs to be multiplied by (n+1)(n+8)  also....then compute....

       Let me know if ya need more help....

Solve (2 + x)^(-1) = x.

It is very likely that your interpretation and that of Melody are the accurate ones. Hence, there are only 34 such numbers as follows:

3001  3031  3061  3091  3121  3151  3181  3211  3241  3271  3301  3331  3361  3391  3421  3451  3481  3511  3541  3571  3601  3631  3661  3691  3721  3751  3781  3811  3841  3871  3901  3931  3961  3991  Total =  34

Which sum up to a total =118,864.

I think that the question is badly worded.

The phrase " when divided by any of the first three primes  ", is that to mean  " if any one of them " or is it to mean " each one of them " ?

I'm with Melody's interpretation, that it means each one of them, (each one of them rather than any single one of them).

That gives rise to the sequence 3001, 3031, 3061, 3091, 3121, ... , 3991.

I don't agree with Melody's method of summation though.

(I get a sum of 118,864.)

\(\frac{1}{x+2}=x\)

x^2+2x=1

x^2+2x-1=0

\(-2\pm \sqrt{4+4}\) / 2

-1+sqrt2

or

-1-sqrt2

I think this is what the questioner might be looking for:

So 4 throws.

The probability of Sanjeet hitting a target is 3/4.  How many minimum number of times must he fire so that the probability of hitting the target of at least is more than 0.99? 

Is something missing?  Maybe  "...the probability of hitting the target of at least one time is more than 0.99?" 

Each individual time Sanjeet fires, his probability of hitting the target is always 0.75 each individual time.

Firing more than once will increase Sanjeet's probability of hitting the target, in regard to the total attempts.

Think about what would be the probability of Sanjeet not ever hitting the target, if he fired, oh say 100 times?

I think if Sanjeet fires twice, the probability of hitting the target is 0.75 + 0.75 = 1.5 

Therefore he has to fire TWO TIMES to have a probability greater than 0.99 ...

note that even though this is a probability, it's still not a certainty. 

I may be all wet here.  A guy can go broke in a hurry betting against Chris. 

_.

If we show that the measure of angle f in the triangle BCF is 90 degrees, then the law of cosines

would tell us that

\(6^2=BF^2+CF^2-2(BF)(CF)cos(f)=BF^2+CF^2-2(BF)(CF)cos (90) \)

\(=BF^2+CF^2\)

since cos(90) = 0.

But f is indeed 90 degrees:

The three interior angles of BFC add up to 180, so

\(f+\alpha +c+\beta+b=180\)

The three interior angles of DBC add up to 180, so

\(\alpha+2b+\beta=80\)

and from the interior angles of EBC we get

\(\beta+2c+\alpha=100\)

Adding the second and third equation we get

\(2\alpha+2\beta+2b+2c=180\), giving us

\(\alpha+\beta+b+c=90\).

This fact added to the first equation results in

\(f=\alpha+\beta+b+c=90\)

Hi Melody: Guest #4 comment is not me.

There are 1,000 numbers between 3000 and 4000. Their average value is 3,500 per number. So, their maximum sum should not exceed:3,500 x 1000 =3,500,000. Your total sum of 50 592 000 / 1 000 =50 592 per number. Hence the comment by Guest #4. 

Take the first 20 numbers: 3001  3003  3004  3005  3006  3007  3009  3010  3011  3013  3015  3016  3017  3019  3021  3022  3023  3025  3026  3027. Each one of the above numbers when tested on mod 2, mod 3 and mod 5 leaves a remainder of 1, hence it must be counted and summed up. At least that is what I understand the question to be. Of course, I could be wrong too.  

Let the pieces be x, y and 5-(x+y)

where 

0

0

also

0<5-(x+y)<5

-5<-(x+y)<0

0

y< -x+5,     y>-x

Draw that up on a number plane and you get a triangle with an area of 12.5  that is your sample space.

Now graph the favourable area.   And work out what it is.

x>1

y>1

5-(x+y)>1

You can take over from here.

The area of trapezoids ABEC, ABFD, and ABFC are 133, 140, and 161 square feet, respectively.

Quadrilateral ABED is a rectangle.

The length of segment DE is 8 feet.

What is the length of segment CF?

\(\text{Let $A_1= ABEC = 133$} \\ \text{Let $A_2= ABFD = 140$} \\ \text{Let $A = ABFC = 161$} \\ \text{Let $A_{\text{rectangle}}=A_\square = 8 AD $} \\ \text{Let $x = CF$} \)

\(\begin{array}{|rcll|} \hline A_1 + A_2 &=& A + A_\square \\ 133+140 &=& 161 + A_\square \\ A_\square &=& 133+140- 161 \\ \mathbf{A_\square} &=& \mathbf{112} \quad | \quad A_\square = 8 AD \\ 8 AD &=& 112 \\ AD &=& \dfrac{112}{8} \\ \mathbf{AD} &=& \mathbf{14} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A &=& \dfrac{(8+x)}{2}*AD \\ 161 &=& \dfrac{(8+x)}{2}*14 \\ 161 &=& (8+x)*7 \\ 23 &=& 8+x \\ x &=& 23-8 \\ \mathbf{x} &=& \mathbf{15} \\ \hline \end{array}\)

The length of segment CF is 15

Quite possibly.  What do you base this assertion on?

If
\(\sin(x) + \cos(x) = \dfrac{1}{2}\),
find
\(\sin^3(x) + \cos^3(x)\).

\(\begin{array}{|rcll|} \hline \Big( \sin(x) + \cos(x) \Big)^2 &=& \sin^2(x) + 2\sin(x)\cos(x)+\cos^2(x) \\ \Big( \sin(x) + \cos(x) \Big)^2 &=& \sin^2(x)+\cos^2(x) + 2\sin(x)\cos(x) \quad | \quad \sin^2(x)+\cos^2(x)=1 \\ \Big( \sin(x) + \cos(x) \Big)^2 &=& 1 + 2\sin(x)\cos(x) \quad | \quad \sin(x) + \cos(x) = \dfrac{1}{2} \\ \Big(\dfrac{1}{2} \Big)^2 &=& 1 + 2\sin(x)\cos(x) \\ 2\sin(x)\cos(x) &=& \dfrac{1}{4} - 1 \\ 2\sin(x)\cos(x) &=& -\dfrac{3}{4} \\ \mathbf{ \sin(x)\cos(x) } &=& \mathbf{-\dfrac{3}{8} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \Big(\sin^2(x) + \cos^2(x)\Big)\Big( \sin(x) + \cos(x) \Big) &=& \sin^3(x) +\sin^2(x)\cos(x) +\cos^2(x)\sin(x) + \cos^3(x) \\ \Big(\sin^2(x) + \cos^2(x)\Big)\Big( \sin(x) + \cos(x) \Big) &=& \sin^3(x)+\cos^3(x) +\sin(x)\cos(x)\Big(\sin(x) +\cos(x)\Big) \\ 1*\dfrac{1}{2} &=& \sin^3(x)+\cos^3(x) -\dfrac{3}{8}*\dfrac{1}{2} \\ \dfrac{1}{2} &=& \sin^3(x)+\cos^3(x) -\dfrac{3}{16} \\ \sin^3(x)+\cos^3(x) &=& \dfrac{3}{16}+\dfrac{1}{2} \\ \mathbf{ \sin^3(x)+\cos^3(x) } &=& \mathbf{ \dfrac{11}{16} } \\ \hline \end{array}\)

You have made a mistake

Find the number of equilateral triangles in the grid (\(n=5\)).

\(a(n) = \Big\lfloor \dfrac{n(n+2)(2n+1)}{8} \Big \rfloor \)

Consider all positive integers between 3000 and 4000 which, when divided by any of the first three primes, leave a remainder of 1. 

Find the sum of all these integers.

(3000+x) mod 2 = 1     therefore  x is odd

(3000+x)mod 3=1         

so

(3000+x) mod6 =1

(3000+x)mod5 =1

so

(3000+x)mod30=1

The smallest number that meets this requirement is   x=1

the next is x=31,

then it goes up by 30.

The numbers will be     

3000(1+31+61+ ........ 991)

Sum=3000*34+(1+(1+30)+(1+2*30)+ ........ (1+33*30))  

         Consider the AP     a=1,  d=30,   n=34, 

\(\qquad S_{n}=\frac{n}{2}(a+L)\\ \qquad S_{34}=\frac{34}{2}(1+991)\\ \qquad S_{34}=17*(992)\\ \qquad S_{34}=16864\\\)

Sum= 3000*34+16864 =   118864    

I have just reapaired my formulas in red.  Now the answers to this interpretation of the question are the same.

Well this is amazing ! While I was writing my solution down, laboring through my meter readings, heureka solved the problem with a neat table, and what do you know he/she arranged the buckets in exactly the same order as I did! There is definitely magic involved here.

Arrange the buckets as follows:

Full 8-gallon bucket--5-gallon bucket---3-gallon bucket---empty 8-gallon bucket

This arrangement is just for ease of reference and involves no magic. The water content of the buckets at the start of the process can be denoted by 8--0--0--0 (Full with 8 gallons--empty--empty--empty). In 7 steps, if I am not dreaming, the first and last buckets will each contain 4 gallons of water; so the water content meter will  read 4--0--0--4.

Step 1: Fill the 5-gallon bucket with the water from the full 8-gallon bucket (meter reading 3--5--0--0).

Step 2: Fill the 3-gallon bucket with the water from the 5-gallon bucket (meter reading 3--2--3--0).

Step 3: Pour the remaining two gallons of water in the 5-gallon bucket into the empty 8-gallon bucket (meter reading 3--0--3--2).

step 4: Empty the water content of the 3-gallon bucket into the 5-gallon bucket (meter reading 3--3--0--2).

Step 5: Fill the 5-gallon bucket with the water from the leftmost 8-gallon bucket (meter reading 1--5--0--2).

Step 6: Fill the 3-gallon bucket from the water contents of the 5-gallon bucket (meter reading 1--2--3--2).

Last step: Empty the contents of the 3-gallon bucket into the leftmost 8-gallon bucket and the contents of the 5-gallon bucket into the rightmost 8-gallon bucket (meter reading 4--0--0--4).

Farmer Jim has an 8 gallon bucket full with water. He has three empty buckets: 3 gallons, 5 gallons and 8 gallons.

How can he get two volumes of water, 4 gallons each, using only the four buckets?

\(\begin{array}{|l|c|c|c|c|} \hline \text{buckets } & 8\ \text{gallons }& 3\ \text{gallons }& 5\ \text{gallons }& 8\ \text{gallons } \\ \hline \text{Start} & 8 & 0& 0& 0 \\ & 5 & 3& 0& 0 \\ & 5 & 0& 3& 0 \\ & 2 & 3& 3& 0 \\ & 2 & 1& 5& 0 \\ & 2 & 0& 5& 1 \\ & 2 & 3& 2& 1 \\ & 4 & 3& 0& 1 \\ \text{End} & 4 & 0& 0& 4 \\ \hline \end{array}\)

Beats me.    Any takers for this one?

It is very remarkable fact that: [Sqrt(2) + 1]^99 is very nearly a WHOLE integer !!!. Here it is calculated to 100 decimal places:

[sqrt(2) + 1]^99 =78486117902932683818009467010929219214.000000000000000000000000000000000000012741106665980664622184682592965611703655250957377780370113573509566385801635499911578533439100076941402033744061215557831686. When you multiply it by its vert small fraction part, it = 1

Two thousand nine hundred and forty unit squares are assembled together to exactly form an 84 by 35 rectangle. One diagonal of the rectangle is drawn. This diagonal intersects the interiors of how many of these unit squares?

a=3;b=0;c=0;d=0;p=0; cycle:n=a*1000+b*100+c*10+d;if(n%2==1 or n%3==1 or d%5==1, goto loop, goto next); loop:printn," ",;p=p+1; next:d++;if(d<10, goto cycle, 0);d=0;c++;if(c<10, goto cycle, 0);d=0;c=0;b++;if(b<10, goto cycle,0);b=0;c=0;d=0;a++;if(a<4, goto cycle,0);print"Total = ",p

OUTPUT:
There are 733 such numbers
Their total sum =2,565,566

P.S. The above code tests each number such as 3001 mod 2, or 3001 mod 3 or 3001 mod 5. If it has a remainder of 1, then it counts it and sums it up.

If you get a full answer can you explain it here please.

You can send me a private message to let me know you have added it.

Otherwise i might not notice.

Thanks :)