After the preschool fee of $330, Cody paid an additional 770 – 330 = $440 for after-school care. Dividing $440 by the hourly rate of $5.50, we find that Cody’s son must have spent 80 hours in after-school care. 

 Let n – 1, n and n + 1 be positive consecutive integers such that (n – 1) ×n × (n + 1) = 16 × [(n – 1) + n + (n + 1)]. Simplifying and solving for n, we have n(n² – 1) = 16 × 3n → n³ – n = 48n → n³ – 49n = 0

=> n(n² – 49) = 0. So, n = 0 or n² − 49 = 0 → (n + 7)(n – 7) = 0 => n = −7 or n = 7. Since n is a positive integer, it follows that the three consecutive numbers are 6, 7 and 8. The difference between their product and sum is 6 × 7 × 8 – (6 + 7 + 8) = 336 – 21 = 315. 

The ratio of the surface area of Wilbur’s plane to the surface area of Orville’s plane is the square of the scale factor of Wilbur’s mini replica of Orville’s plane. Since Wilbur’s mini replica has linear dimensions that are \(\dfrac{1}{2}\) the size of Orville’s model airplane, it follows that the ratio between the lift forces on Wilbur’s and Orville’s planes is \(\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4}\) 

If we add the wins of both Flying Turtles and Dolphins, we have to then subtract the 19 wins for the games they played against each other  regardless of who won those games. The value of F + D, then, is 95 + 84 – 19 = 160. 

Let x, y be the length and breadth of a rectangle and 16 is the length of diagonal as shown below:

Using the Pythagorean Theorem in the right triangle formed by sides with the diagonal, we get: x² + y² = 256²

=> \(y=\sqrt{256-x^2}\) and \(A\left(x\right)=x\sqrt{256-x^2}\)

that is the function we have to maximize. Notice that x can vary between x = 0 (the rectangle collapses to a vertical line) and x = 16 (the rectangle collapses to a horizontal line), and in both extremes the area is 0. Let’s find critical points for A:

\(0=A'\left(x\right)=\sqrt{256-x^2}-x-\dfrac{x}{\sqrt{256-x^2}}\\ \Rightarrow\dfrac{x^2}{\sqrt{256-x^2}}=\sqrt{16-x^2}\\ \Rightarrow x^2=256-x^2\)

and so \(x=\sqrt{128}\) is the only critical value

Moreover, A(\(\sqrt{8}\)) = 8 > 8 so it is the maximum

A.BC + D.EF = 0.13 + 0.24 = 0.37

After the first spin, the probability that the pointer lands on the same color is: 1 : 3 = 1/3

We have: 110011₂ = 1 x 2⁵ + 1 x 2⁴ + 1 x 2¹ + 1 x 2⁰ = 51₁₀ =  6 x 8¹+ 3 x 8⁰ = 63₈

So 110011₂ is rewritten as 63₈ in octal (base 8)

It is not wrong. Just because of your OLD phone/computer

An amicable (or friend) pair  (m,n)  consists of two integers  m,n  for which the sum of proper divisors (the divisors excluding the number itself) of one number equals the other. Amicable pairs are occasionally called friendly pairs (Hoffman 1998, p. 45), although this nomenclature is to be discouraged since the numbers more commonly known as friendly pairs are defined by a different, albeit related, criterion. Symbolically, amicable pairs satisfy

s_(m) = n
(1)
s_(n) = m
(2)

where

s_(n)=sigma_(n)-n
(3)

is the restricted divisor function. Equivalently, an amicable pair  [(m,n)]  satisfies

sigma_(m)=sigma_(n)=s_(m)+s_(n)=m+n
(4)

where  sigma_(n)  is the divisor function. The smallest amicable pair is (220, 284) which has factorizations

220 = 11·5·2²
(5)
284 = 71·2²
(6)

giving restricted divisor functions

s₍₂₂₀₎ = sum{1,2,4,5,10,11,20,22,44,55,110}
(7)= 284
(8)
s₍₂₈₄₎ = sum{1,2,4,71,142}
(9)
= 220.
(10)

The quantity

sigma_(m)=sigma_(n)=s_(m)+s_(n),
(11)

in this case,  220+284=504 , is called the pair sum. The first few amicable pairs are (220, 284), (1184, 1210), (2620, 2924) (5020, 5564), (6232, 6368), (10744, 10856), (12285, 14595), (17296, 18416), (63020, 76084), ... (OEIS A002025 and A002046). An exhaustive tabulation is maintained by D. Moews.

Still some more, but with "x", "/", ...

123-45-67+89=100

1 + 2 + 3 – 4 + 5 + 6 + 78 + 9 = 100
123 +45 - 67 + 8 - 9 = 100

123 + 4- 5 + 67 - 89 = 100
 

An amicable pair  [(m,n)]  consists of two integers  [m,n]  for which the sum of proper divisors (the divisors excluding the number itself) of one number equals the other. Amicable pairs are occasionally called friendly pairs (Hoffman 1998, p. 45), although this nomenclature is to be discouraged since the numbers more commonly known as friendly pairs are defined by a different, albeit related, criterion. Symbolically, amicable pairs satisfy

[s(m)] [=] [n]
(1)
[s(n)] [=] [m,]
(2)

where

[s(n)=sigma(n)-n]
(3)

is the restricted divisor function. Equivalently, an amicable pair  [(m,n)]  satisfies

[sigma(m)=sigma(n)=s(m)+s(n)=m+n,]
(4)

where  [sigma(n)]  is the divisor function. The smallest amicable pair is (220, 284) which has factorizations

[220] [=] [11·5·2^2]
(5)
[284] [=] [71·2^2]
(6)

giving restricted divisor functions

[s(220)] [=] [sum{1,2,4,5,10,11,20,22,44,55,110}]
(7)
[=] [284]
(8)
[s(284)] [=] [sum{1,2,4,71,142}]
(9)
[=] [220.]
(10)

The quantity

[sigma(m)=sigma(n)=s(m)+s(n),]
(11)

in this case,  [220+284=504] , is called the pair sum. The first few amicable pairs are (220, 284), (1184, 1210), (2620, 2924) (5020, 5564), (6232, 6368), (10744, 10856), (12285, 14595), (17296, 18416), (63020, 76084), ... (OEIS A002025 and A002046). An exhaustive tabulation is maintained by D. Moews.

Why don't you post some?

An amicable pair  [(m,n)]  consists of two integers  [m,n]  for which the sum of proper divisors (the divisors excluding the number itself) of one number equals the other. Amicable pairs are occasionally called friendly pairs (Hoffman 1998, p. 45), although this nomenclature is to be discouraged since the numbers more commonly known as friendly pairs are defined by a different, albeit related, criterion. Symbolically, amicable pairs satisfy

[s(m)] [=] [n]
(1)
[s(n)] [=] [m,]
(2)

where

[s(n)=sigma(n)-n]
(3)

is the restricted divisor function. Equivalently, an amicable pair  [(m,n)]  satisfies

[sigma(m)=sigma(n)=s(m)+s(n)=m+n,]
(4)

where  [sigma(n)]  is the divisor function. The smallest amicable pair is (220, 284) which has factorizations

[220] [=] [11·5·2^2]
(5)
[284] [=] [71·2^2]
(6)

giving restricted divisor functions

[s(220)] [=] [sum{1,2,4,5,10,11,20,22,44,55,110}]
(7)
[=] [284]
(8)
[s(284)] [=] [sum{1,2,4,71,142}]
(9)
[=] [220.]
(10)

The quantity

[sigma(m)=sigma(n)=s(m)+s(n),]
(11)

in this case,  [220+284=504] , is called the pair sum. The first few amicable pairs are (220, 284), (1184, 1210), (2620, 2924) (5020, 5564), (6232, 6368), (10744, 10856), (12285, 14595), (17296, 18416), (63020, 76084), ... (OEIS A002025 and A002046). An exhaustive tabulation is maintained by D. Moews.

Your answer is incorrect, try again

SORRY, the answer is incorrect, but I have made a mistake when for him

You can see the answer at: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/15puzzle.pdf

P/s: You always type incorrectly

Ex: Standand, not stardard

The probablity of rolling one first time is \(\dfrac{1}{6}\)

Six times that, we have: \(\left(\dfrac{1}{6}\right)^6\)=\(\dfrac{1}{46656}\)

But is this correct, because it is BEST IQ, right?

One cubic yard of topsoil is 3 x 3 x 3 = 27 ft² of topsoil. 

The garden is 10 x 8, which is 80 ft², so the depth of the soil will be \(\dfrac{27}{80}\) feet, which is equivalent to \(\dfrac{27}{80}\times12=4\dfrac{1}{20}\) inches

Sorry for OLD answer

First ask a simpler question: What is the maximum number of points of intersection for a line and a square? Because the square is convex, the answer is 2.

A Triangle consists of 3 line segments, each of which is a subset of a line. So each segment can clearly not intersect the square more than twice. This gives 6 as the upper bound, even if you extend each edge of the triangle into an infinite line.

Further note that if both endpoints of an edge are inside the square, then the whole edge is inside the square and doesn’t intersect it at all. And if one endpoint is inside and one is outside, then there can only be one point of intersection.

Therefore, you can only get 6 points of intersection if all 3 corners of the triangle are outside the square. If one is inside, the max is 4; if two are inside, the max is 2; and if 3 are inside, the max is 0.

Of course change the door. What is in your mind???????

Everyday (starting from question 1) I will post 1 - 2 question(s) to get a point:

- The fastest person who answer correctly will get 1 point/question/answer

- You can still answer my old question at https://e-learning.codienhanoi.edu.vn/discuss/member/leanhduy0206 and get point

THE QUESTIONS TODAY ARE:

12) Given \(\Delta ABC\) (AB = AC). Find\(\widehat{ABC}\) 

13) Andy and Beth is in a race at the pool. Andy runs around the pool while Beth swims by the length of the pool. Knowing that:

     - Andy is three times as fast as Beth

     - The time Beth swims 5 turns is equal to the time that Andy runs 6 turns around the pool

     - The length of the pool is 50m

What is the width of the pool?

We might start by examining the number of ways that ONE SIDE of a triangle can intersect a square. 
In other words, in how many ways can a LINE intersect a square?
After a bit of mental imagery, we might conclude that a SINGLE LINE can intersect a square in at MOST 2 ways

A triangle is composed of THREE LINE SEGMENTS.
If each SINGLE LINE can intersect a square in at MOST 2 ways, then the 3-sided triangle can intersect a square in AT MOST 6 ways (with 2 intersections per line) 
So, the correct answer must be 6 or less

At that point, if we're able to sketch a scenario in which there are 6 intersections, we can be certain that this is, indeed, the GREATEST number of intersections. 
 

Call Amos is A, Butch is B and Cody is 3

C_1:  - If A is the one to shoot first and shoot C, he will be killed by B

     => SO A will shoot B first

     Then it's time for C to shoot A

     => The percentage of A alive is 50% and C is always alive (out of shot)

 C_2: - If B is the one to shoot first and shoot C, he will be killed by A

     => SO B will shoot A first

     Then it's time for C to shoot B

     => The percentage of B alive is 50% and C is always alive (out of shot)

 - If C is the one to shoot first 

     + He shoots A:

          C₃. If shoot to target, B will kill C => B is alive

          C₄. If shoot miss, B will kill A or A will kill B (reason above) and then C and A (or B) are alive (out of shot)

  + He shoots B:

          C₅. If shoot to target, A will kill C

          C₆. If shoot miss, B will kill A or A will kill B (reason above) and then C and A (or B) are alive (out of shot)

SO WE HAVE:

          ALIVE        50%        DEAD

C₁:         C               A               B

C₂:         C               B               A

C_3:           B                              A , C

C_4:       C, A (or B)                B (or A)

C₅:         A                               B, C             

C₆:       C, A (or B)                B (or A)         

SO C HAS THE BIGGEST CHANGE TO LIVE (\(\dfrac{4}{6}=66.66\%\) )

     A (or B)' s probability to live;\(\dfrac{1-\dfrac{4}{6}}{2}=\dfrac{1}{6}=16.66\%\)

Of the 18 two-digit multiples of 5 from 10 to 95, only 13 have exactly distinct prime numbers.

=> The probability of choosing one of these 13 at random, is \(\dfrac{13}{18}\)

Everyday (starting from question 1) I will post 1 - 2 question(s) to get a point:

- The fastest person who answer correctly will get 1 point/question/answer

- You can still answer my old question at https://e-learning.codienhanoi.edu.vn/discuss/member/leanhduy0206 and get point

THE QUESTION TODAY IS:

11) Mr. Cook bought a big box of his favourite Halloween candies to hand out to trick-or-treaters. However,
he ate half of all the candies himself before the first child came and took her share. He ate half of what
was left before the second child came, and half of what was left again before the third child came and
took all of the remaining candies. If it is known that each child received the same number of candies,
which of the following statements is certainly true about how many candies Mr. Cook originally bought?
(A) It is a multiple of 3

(B) It is a multiple of 4

(C) It is a multiple of 6

(D) It is a multiple of 7

(E) It is a multiple of 11

(Full answer please)