All Answers 358087 Answers

The limit of the argument is zero, but that doesn't make the sum zero.

2^3 + 3^1 = 11, 2^3 + 3^6 = 737 = 11*67, 2^3 + 3^11 = 177155 = 16105*11.

2^13 + 3^1 is divisible by 11, same as 2^23 + 3^1 is also. Along with 2^13 + 3^6...

As you will notice with these calculations (that you can do brutally by hand or easily with a calculator (to find patterns, of course)), the values of m and n for 2^m + 3^n is divisible by 11 is m = 10a + 3, n = 5b + 1.

List for m: 3, 13, 23, ..., 593 = 60 values

List for n: 1, 6, 11, ..., 596 = 120 values

In total, there are 60 * 120 = 7,200 correct cases, along with 600 * 600 = 360,000 total possible cases: coming to a probability of 7200/360000 = 72/3600 = 2/100 = 1/50, or 2% chance.

Now, I leave you with this parmen... how can you prove that the generalizations for the patterns of m and n are correct?? 

We can do complimentary counting, where you have the total choices (1) minus the wrong choices.

The only way that the machine can generate four numbers where the product is NOT divisible by 2 is if and only if they are all odd numbers. From 1 through 9, there are 5 odd and 4 even numbers.

For the first number generated, there is a 5/9 chance it is odd, and for the next number 5/9, the third 5/9, and the last 5/9 chance again.

Thus, the probability is \({5^4\over{9^4}}={625\over{6561}}\) for generating four numbers with a product not divisible by 2. 

To calculate the probability that the machine will generate four numbers where their product is divisible by 2, you just subtract 625/6561 from 1, which equals to = \(5936/6561\) probability.

This one is pretty easy

The triangle is equilateral

The altitude of an equilateral triangle  is a median

And since M is the midpoint of ST, then UM is also a median

Their intersection, X, is the centroid of STU

SN = sqrt [ST^2 - TN^2]  = sqrt [ 8^2 - 4^2]  = sqrt [ 48]  = 4sqrt (3)

By a property of centroids ....SX = (2/3)(SN)

So

SX = (2/3) * 4 (sqrt3)  = 8sqrt (3)  / 3  =  8 / sqrt 3   ≈  4.618

Scroll down to see answer, very long.

I'll give this my best attempt 

We see that without the condition, the number of total possibilities of arranging the indistinguishable fruits is \(\frac{8!}{4!4!}=70\).

Now add this condition in, and we need to find the total number of cases that don't satisfy this condition, or otherwise that there is at least one instance of an apple that is surrounded by apples.

Lets split that into 3 sub-conditions 

Condition 1) An apple is first and the second fruit is also an apple

Condition 2) An apple is last and the second to last fruit is also an apple

Condition 3) An apple is not first or last, but the two fruits next to it are apples. (3 apples in a row)

Now we have lots of cases to consider.

Satisfying condition 1, we have \(\frac{6!}{4!2!}=15\) cases for condition 1.

Satisfying condition 2, we also have \(\frac{6!}{4!2!}=15\) cases for condition 2.

Condition 3 is more difficult, because 3 apples in a row, also includes 4 apples in a row, but we overcount, the cases for 4 apples in a row.

First we count the cases for 3 apples which, we treat 3 apples a group, or a seperate variable, so we order the group, one apple, and 4 pears getting \(\frac{6!}{4!1!1!}=30\), but we overcount many cases with 4 apples in a row. Each case with 4 apples in a row can be formed 2 ways, for example consider the sequence PAAAAPPP where P is pear and A is apple. This sequence can be formed by P + AAA +APPP or also PA + AAA+PPP with our counting, so we count each 4 in a row case twice, so we need to subtract once the cases with 4 in a row.

Again, to count the number of 4 in a row cases, we treat the 4 as a group, a seperate item, so there is a total of \(\frac{5!}{4!1!} = 5\) 4 in a row cases. Subtracting this, we get \(30-5=25\) total cases for condition 3.

Now we count the overlaps. We see that there is 1 case of condition 1 and condition 2, AAOOOOAA.

We see there is 5 cases of condition 1 and condition 3. AAA + an A in any of the 5 remaining spaces, yields 5 cases.

We see there is also 5 cases of condition 1 and condition 3. an A in any of the first 5 spaces + AAA, yields an additional 5 cases.

We see there is not possible a case of condition 1 and condition 2 and condition 3, because after AAOOOOAA we can't satisfy condition 3.

So now we have. \(15+15+25-5-5-1=44\)cases of apples being surrounded by apples. 

This is the exact opposite of the case we want to find, so we do total cases - cases of apples being surrounded by apples for a total of \(70-44=26\) cases. Our final answer is 26 arrangements.

Oscar and Felix leave the same house driving separately to the same conference, along the same route, at a speed of $x$ mi/h. If Oscar had driven $10$ mi/h faster, he would have driven for $15$ fewer minutes. If Felix had driven $30$ mi/h faster, he would have driven for $20$ fewer minutes. What is the distance from their house to the conference?

Let D be the distance , x be the normal rate  and  T  be the normal time (in hours)

15min  =1/4 hr     20 min = 1/3 hr

We have this system

D / x    =  T     →   D = Tx  (1)

D / ( x + 10)  = T - 1/4     (2)

D / ( x + 30)  = T - 1/3     (3)

Sub (1) into (2) , (3)

Tx / (x + 10) = T -1/4

Tx / ( x + 30) = T -1/3

Tx  = (T -1/4)(x + 10)

Tx  = (T -1/3)(x + 30)

Tx = Tx - (1/4)x + 10T - 5/2

Tx = Tx -  (1/3) x  + 30T -10

(1/4)x - 10T  = -5/2  →   x - 40T   =  -10      (4)

(1/3)x - 30T = -10  →    x - 90 T  =  -30       (5)     subtract (5) from (4)

50T = 20

T = 2/5 hrs

x =  40T - 10  =  40 (2/5)   - 10   =   6 miles/hr

The distance is   (2/5 - 1/4) (6 + 10)  =  (3/20)(16)  = 48/20  = 24/10  = 2.4 miles

Let  Q = (0,0)

QN = 1      RN = (PR/2)  = 1/2  = PR

Triangle QNR is right with QNR the right angle

Then let  QR  =  sqrt [ QR^2 - RN^2 ] =   sqrt [ 1 - (1/2)^2 ] =  sqrt (3/2)

So  R = (sqrt (3) / 2,  0)

Also  QP = sqrt [ QR^2  - PR^2 ]   = sqrt [ 1 -(1/2)^2] = sqrt (3/2)

Now  we can find the coordinates  of   P  by the intersection of two circles

One centered at (0,0) wih a radius of sqrt (3)/ 2

And one centered at R = (sqrt (3) / 2  , 0)  with a radius of 1

The equations are

x^2 + y^2   = 3/4    →   y^2 = (3/4) - x^2

(x - sqrt (3)/2)^2  + y^2   = 1 → y^2 = 1 - (x - sqrt (3)/2)^2

Setting these equal  and  solving for the x coordinate  of P

(3/4) - x^2  =  1  -  (x - sqrt (3)/2 / 2)^2 

(3/4) - x^2  = 1 - x^2 + sqrt (3)x - (3/4)

1/2 =  sqrt (3) x

x =  1/ (2sqrt 3)  =  sqrt (3)  / 6

And y^2  = (3/4) -  3  / 36   =   3/4 - 1/12  = 32/48  = 2/3

y = sqrt (2/3)

So  P  = ( sqrt (3) / 6  ,  sqrt (2/3))

Since N is a  median of QP......then N  =  (sqrt (3) /12 ,  sqrt (2/3) / 2 )

O is the intersection of medians QN and RN  { O is the centroid of PQR)

By a property of centroids.... ON   is  (2/3)  the distance from  R to N

(2/3)  sqrt  [   (sqrt (3)/2  - sqrt (3)/12)^2  + ( sqrt (2/3) /2)^2  ]   

(2/3)  sqrt [ (5sqrt (3)/12)^2  +  ( sqrt (2/3)  /2 )^2 ] =

(2/3) sqrt [ 75/144 + 1/6 ]   = 

(2/3) sqrt  [75 /144 + 24/144 ]  =

(2/3) sqrt (99 / 144)   =

(2/3) (1/12) sqrt (99)   =

3 (2/3) (1/12) sqrt (11) =

sqrt (11) / 6

Hi LioLaeus504! 

By definition, a TANGENT is a line touching circles at one point that doesn't intersect through the circle (which would result in 2 points). That being said, a common external tangent cannot be through the circles, and thus cannot be the connections of the two centers...

And for your question, what it is asking is that imagine that PQ is extended infinitely to the right, and the external tangent (on top of the two circles) at some point intersects this line PQ at a point R. Continue from there and best of luck! 

Clever solution hairy! If you don't know trig, this is the way to go- 

The area of a triangle is \(ab\sin{C}\over{2}\), along with b*h*1/2, but I prefer the first one for this problem. (You can look the proof up, it is pretty simple)

We know that a = b = 7, so the area of the triangle = 49sinC/2 = 14. Then solving for sinC = 4/7.

Because we want to find the third length, it is best to use the Law of Cosines for C, but how?

Using the identity \(\sin^2\theta+\cos^2\theta=1\), where \(\theta=C\), we know that \(\cos^2C=1-\sin^2C=1-({4\over7})^2=1-{16\over{49}}={33\over{49}}\). Square rooting both sides, you get \(\cos{C}=\pm{\sqrt{33}\over{7}}\). Now, we can effectively do Law of Cosines: c^2 = a^2 + b^2 - 2ab*cosC.

We will do 2 seperate calculations of c from law of Cosines based on the two possible values of cosC.

Case 1: \(\cos{C} = +{\sqrt{33}\over{7}}\), keep in mind that c is the third side length we don't know, and a = b = 7.

\({c_1}^2=7^2+7^2-2*7*7*{\sqrt{33}\over{7}}=98-14\sqrt{33}\), and since c has to be positive to be a real triangle:

\(c_1 = \sqrt{98-14\sqrt{33}}\)

Case 2: \(\cos{C}=-{\sqrt{33}\over{7}}\); a = b = 7 still.

\({c_2}^2=7^2+7^2-2*7*7*-{\sqrt{33}\over{7}} = 98 + 14\sqrt{33}\), which yields

\(c_2=\sqrt{98+14\sqrt{33}}\), (again, there is no plus/minus because c has to be positive to be a real triangle.)

Setting c1 = a + b*sqrt(c) is a great strategy for radical simplification, and then you can square both sides and solve system of equations (with substitution, and yes, I did it)! Oh and by the way, don't expect a good answer for this problem...

After simplification, you should reach \(c_1=\sqrt{77}-\sqrt{21}; c_2=\sqrt{77}+\sqrt{21}\). 

Product of 2 possible perimeters = (7 + 7 + c1)(7 + 7 + c2) = [14 + sqrt(77) - sqrt(21)][14 + sqrt(77) + sqrt(21)], and you can do difference of squares: a^2 - b^2 = (a + b)(a - b), where a = 14 + sqrt(77) here, and b = sqrt(21):

= 252 + 28sqrt(77)

Hence, our final answer is \(252 + 28\sqrt{77}\). 

Nice!

I loved how you included the units in your calculation. That helps the understanding of the problem a lot. 

A Nice trick, that some people don't know about, is that you can just directly multiply the units. 

For in example \(3 hour * \frac{2 miles}{hour} = 6 miles\), the hours cancel out. Just remember to multiply by the correct unit fraction, because \(3 hour * \frac{hour}{2miles}=\frac{3{hour}^2}{2miles}\), and while equivalent, not our desired unit. This can done multiple times to get any desired unit, which can be very useful in rate problems.

Happy Solving!

We see trigonmetry might be a possible solution to the problem, but here is a non-trigonmetry solution.

Instead, we focus on the fact that this triangle is an isosceles triangle, so we can draw a perpendicular to solve this problem.

Consider the diagram below.

We can now use a system of equations to solve this problem. \(\sqrt{{x}^{2}+{y}^{2}}=7, \frac{2xy}{2}=14\)

Simplifying, we get \({x}^{2}+{y}^{2}=49, xy=14\). This is not an easy system to solve, but there is a small trick I often use for a problem like this. We recognize the structure of \(x+y=\sqrt{{x}^{2}+2xy+{y}^{2}}=\sqrt{49+14*2}=\sqrt{77}\). Now we can use a little trick with the Vieta's formulas. We recognize that x and y are the two roots of the equation \({a}^{2}-\sqrt{77}a+14\). (because the sum of the roots is sqrt(77) and the product of the roots is 14). We recoginze here, that x and y are interchangable, because we can assign any of the roots to x or y.

But we need to read the problem. It is not asking the roots, but the product of the perimeters. Because we know x and y are the two roots, or the two solutions to the base, we calculate \((14+2x)(14+2y)=196+28(x+y)+4xy=196+28\sqrt{77}+56=252+28\sqrt{77}\). So our answer is 252+28sqrt(77).

Note the basic formula: st = d, speed * time = distance

So, using our statements in the question and the above formula, we can formulate these 3 equations:

x mi/hour * t hours= d miles

(x + 10)mi/hour * (t - 1/4)hours = d miles

(x + 30)mi/hour * (t - 1/3)hours = d miles

You have a system of 3 variables and 3 equations, which will be solvable as long as there is no redundancy (which there isn't).

I trust that you can continue the problem from here- as formulating equations are my nightmare for rate problems... good luck! 

Thanks Imcool and heureka- I just realized that one could use triangle inequality then manipulation here too! (BTW, the answer is n)

Here : https://web2.0calc.com/questions/right-triangle_44

ab + 3a + 8b

a     b       n

11    2     71

 4     5     72

6     4      74

3     6      75

2     7      76

9     3      78

5     5      80

7     4      81

4     6      84

3     7      86

6     5      88

11   3      90

\(h^2=1^2-(2r-2)^2=r^2-(2-r)^2\\ 1-(4r^2-8r+4)=r^2-(4-4r+r^2)\\ 1-4r^2+8r-4=r^2-4+4r-r^2\\ 4r^2-4r-1=0\\ r^2-r-0.25=0\\ \)

\(r=0.5\pm\sqrt{0.25+0.25}\\ r\in\{1.207,-0.207\}\\ \color{blue}r=0.5+\sqrt{0.5}=1.207\\ h^2=1^2-(2r-2)^2\\ h^2=1-(1+\sqrt{2}-2)^2\\ h^2=1-(\sqrt{2}-1)^2\\ h^2=1-(2-2\sqrt{2}+1)\\ h=\sqrt{2\sqrt{2}-2}\\ \color{blue}h=\overline{CH}=0.91\)

!

n  = 12

To see why

12!   =

12  * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2

Regrouping,  we have

12 * (8 * 6) * (4 * 3) * 9 * 11 * 10 * 7 * 5 *  2  = 

12 * (12 *4)  *  (4 * 3) *  (9) * 11 * 10 * 7 * 5 * 2 =

12 * (12) * (4 *3) * ( 9 *4) * 11 * 10 * 7 * 5 *2  =

12 * 12 * 12 * (12 *  3) * 11 * 10 * 7 * 5 * 2 =

12 * 12 * 12 * 12 * 3 * 11 * 10 * 7 * 5 * 2

12^4 * 23100

So

12!  /12^4  = 

12^4 * 23100  / 12^4   =   23100

We can calculate a generating polynomial   by the sum  of differences

12     44      100      186     308

     32       56     86        122

          24      30      36

                 6       6

                      0

We have 3 rows of non-zero differences

So we can generate a  3rd power polynomial  for the nth  group sum

an^3 + bn^2  + cn + d         where n is the nth group

We have  this system

a + b + c + d   = 12

8a + 4b + 2c + d = 44

27a + 9b + 3c + d  = 100

64a + 16b + 4c + d = 186

Solving this gives us

a =1

b= 6

c= 7

d= -2

The generating polynomial for the sum of the nth group is :

n^3 + 6n^2 + 7n  - 2

So.....the sum of the 5th group  is

5^3 + 6*5^2 + 7*5 - 2    =   308

We can find three infinite sums and add these fractions to 38 /1000

(1*10^(-4)) /  ( 1 -10^(-3) )  = 1/9990

(4*10^(-5)) / (1 -10^(-3)) = 1/24975

(6*10^(-6)) / (1-10^(-3))  = 1/166500

38/1000 + 1/9990 + 1/24975 + 1/166500  =  9527  / 249750