All Answers 356035 Answers

1 apple + 1 apple is 2 apples

A dice is rolled twice. What is the probability of observing at least one five?

P(first one 5 )+P(both 5) - P(both 5)

I'm going to try doing it a couple of different ways and see I get the same answers :)

\(\frac{1}{6}+\frac{1}{6}-\frac{1}{6}*\frac{1}{6}\\=\frac{12}{66}-\frac{1}{36}\\=\frac{11}{36}\\~\\ or\\ \frac{1}{6}*\frac{5}{6}+\frac{5}{6}*\frac{1}{6}+\frac{1}{6}*\frac{1}{6}\\ =\frac{5}{36}+\frac{5}{36}+\frac{1}{6}\\=\frac{11}{36}\)

7x – 3(x + 2) = 2(x + 4)

\(7x – 3(x + 2) = 2(x + 4)\\ 7x + \;\;\;– 3(x + 2) = 2(x + 4)\\ 7x + \;\;\;– 3*x + -3*2 = 2*x +2* 4\\ 7x + \;\;\;– 3x + -6 = 2x +8\\ \qquad 7x -3x -6 = 2x +8\\ \qquad \qquad 4x -6 = 2x +8\\ \qquad 4x -6-2x = 2x +8-2x\\ \qquad \qquad \;\;2x -6 = 8\\ \qquad \quad 2x -6+6 = 8+6\\ \qquad \qquad \qquad \;\;2x = 14\\ \qquad \qquad \qquad \;\;\;x = 7\\\)

checked

LHS=49-3(9)=49-27=22

RHS=2(11)=22=LHS     excellent :))

Print and stick together.

Tree Diagram

1 + 13/100   is just    1 13/100  

= 1+13/100 = 1.13

Quotient =  5 /(-4+(-2))       OR      (-4+(-2))/5   since the question did not specify which was divisor and which was dividend.

Wow. Any other methods? My classmates do not know calculus!!

Each time you roll the die (one is a 'die'  ...two are called 'dice') you have a 1/6 chance of rolling a '5'

rolling TWO times DOUBLES that chance    2 x 1/6 = 1/3 chance of rolling a '5'

3x4+6-x=5-3

12 + 6 - x = 2

18 -x = 2

18 = 2+x

16 = x

2 (a-b)   does not equal  (a-b)^2   is your wrong turn

(b^2-2ab+a^2) / 2(a-b) =

(b-a)^2 / (2(a-b) =

(b-a)^2 /(-2(b-a)) =

(b-a) / -2 =

(a-b)/2

1/2 (a-b)

2 parts + 4 parts + 5 parts + 1 part = 12 parts total of which 4 parts are sand   

4 out of 12   or 4/12 sand  out of 6.5 cubic meters     4/12 x 6.5 =

4/12*6.5 = 2.167 cubic meters of sand

use a calculator u idot

is it thelegend27?

The result  of this summation / subtraction is the same as the solution for the sllghtly more advanced equation found here :

http://web2.0calc.com/questions/how-to-do-this_699

p  = k / (q + 4)^2

And we know that

2  = k / (2 + 4)^2

2  = k / (6)^2

2 = k / 36

2(36)  = k  = 72  = the proportionality constant

So

p = 72 / ( -2 + 4)^2

p = 72 / (2)^2

p  = 72 / 4

p = 18     when q  = -2

Cost price + .20* Cost price  = $20.40

Cost price ( 1 + .20)  = $20.40     divide both sides by (1 + .20)  = (1.20)

Cost price  = $20.40 / 1.20  = $ 17

The answer is 42.....or something like that

But 8 times 4 is 32..not 36??

MaxWong

I think we should use Euclidean Algorithm for that.

Wrong, Max......

GA  applied the [ lesser known] "Lancelot Link Falling Off A LOG Algorithm"  to solve this.....

Those chimps of his don't monkey around.......

P.S -  Rumor has it that  Sir LaTex  also provided some technical assistance.........

18. Figure: http://imgur.com/a/euSF9

In the figure, ABCD is a rhombus. E is a point on BC such that AE⊥BC, cos B = 4/5 and EC = 2. P is a point on AB. Find the least value of PE + PC.

I worked my  way through this....[ well, almost ].....I believe that P minimizes PE + PC  when

P = (256/45 , 64/15  )

To start, we need to find the coordinates of E  and C.....let B  = (0, 0)

And since cosB = 4/5, then BE / AB  = 4/5  → BE  = (4/5)AB  and EC = 2

And BC  = AB....so

(4/5)AB + 2  = AB

2  = (1/5)AB  → AB  =10 = BC......so BE  = 8

So  E  = (8,0)   and  C  = (10,0)....and ABE  is a 6, 8, 10 right triangle, with A = (8,6)....therefore, the tanB  = 6/8  = 3/4....and for any point x on P, y =(3/4)x

So......we have this

D  = PE + PC

D = √[ (x - 8)^2  + [(3/4)x]^2 ]  + √[ (x - 10)^2  + [(3/4)x]^2 ]

D = √[ (x^2 - 16x  + 64  + (9/16)x^2 ]  + √[ (x^2 - 20x  + 100  + (9/16)x^2 ] 

D = √[ (25/16)x^2 - 16x  + 64  ] + √[ (25/16)x^2 - 20x  + 100  ]

So.....taking the derivative, we have

D'  =  [ (25/8)x - 16] /  √[ (25/16)x^2 - 16x  + 64  ]  + [ (25/8)x - 20] /  √[ (25/16)x^2 - 20x  + 100  ] 

Set this = 0  and solve for x...[ I let WolframAlpha do the heavy lifting here.....if I have time  tomorrow, I might  see if I can go through the messy Algebra associated with this....but....I'm not promising anything.........I know you're not supposed to use CAS help....but I'm not in the competition....!!! ]

Anyway....setting to 0 and solving for x  should result  in 

P = [x, (3/4)x ]  = (256/45 , 64/15  )

See post #8

19)  Solution for Greatest Common factor of ((3^2046) - 1) and ((3^2016) - 1)

General formula

\(\gcd(b^m-1,b^n-1)= (b^{\gcd(m,n)}-1). \\ \gcd(3^{2046} - 1, 3^{2016}-1)= (3^{\gcd(2046,\;2016)}-1)\\\)

\(3^{2046} = 2046*log(3)\ \& \ 3^{2016} = 2016*log(3) \text { |Note that exponents become factors: } \\ \gcd (2046), (2016) = 6 \text {| Find the GCF of the exponents. }\\\)

\((3^{6}) -1 \text {| Use this factor as an exponent of the base and subtract the one. }\\ (3^{6}) -1 = 728 \text { |Solve in base ten. }\\\) \(\text {Descriptive mathematics for the example. }\\ (3^{2046}- 1) = ((3^{6})-1)((3^{2040})+(3^{2034})+(3^{2028})+(3^{2022})+ (3^{2016}) . . . +(3^{6}) + 1)\\ (3^{2016} - 1) = ((3^{6})-1)((3^{2010})+ (3^{2004})+(3^{1998})+(3^{1992})+ (3^{1988}) . . . +(3^{6}) + 1)\\ \)

(My mentor – Lancelot Link, taught this method to me.)

holy cow 

(-8)(-9) - 7