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???????

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 \(\dfrac{13}{18}\)

The total probability of rolling two dices is: 6 x 6 = 36

We have:

3 + 6 = 4 + 5 = 5 + 4 = 6 + 3 = 9

4 + 6 = 5 + 5 = 6 + 4 = 10

5 + 6 = 6 + 5 = 11

6 + 6 = 12

=> The total probability that values on the 2 tops faces add to at least 9 is: 4 + 3 + 2 + 1= 10 (probability)

Hence, the answer is \(\dfrac{10}{36}=\dfrac{5}{18}\)

     A diameter of 9 inches is \(\dfrac{3}{4}\) of one foot. 

     We want to know how many \(\dfrac{3}{4}\) of one foot there are in 12 feet. Dividing, we get \(12:\dfrac{3}{4}=16\)

     So, to create the 12-foot wall, Gerald will have to stack 16 logs

     At 5:42, the time elapsed is 42 of the 60 minutes in the 5 o'clock  hour. Since the minute hand will make the compele revolution during that hour, at 5:42 it has traveled \(\dfrac{42}{60}=\dfrac{7}{10}\) of the full 360 degrees, or\(\dfrac{7}{10}.360^o=252^o\).

     The hour hand makes \(\dfrac{1}{12}\) of a complete revolution every hour. So, from 12:00 to 5:00, it travels \(\dfrac{5}{12}\) of the full 360 degrees, or \(\dfrac{5}{12}.360^o=150^o\)

     By 5:42, it has traveled another \(\dfrac{7}{10}\) of \(\dfrac{1}{12}\) of the full degrees, or \(\dfrac{7}{10}.\dfrac{1}{12}=21^o\), for a total of 150 + 21 = 171 degrees.

     The measure of the angle between the hands at 5;42 is 252 - 171 = 81 degree

\(ANSWER:81^o\)

Since 2018=2⋅1009, where both 2 and 1009 are primes, so 2018 has factors 1, 2, 1009 and 2018, so D(2018)=4σ0(2018)=4.

 (Divisor function - Wikipedia)

Maura's present age = x

Cara's present age = y

=> x + 5 = y => (x - 7) + 5 = y - 7

     (x - 7) = \(\dfrac{1}{2}\)(y - 7)

=>  \(\dfrac{1}{2}\)(y - 7) + 5 = y - 7 => \(\dfrac{1}{2}\)(y - 7) = 5 => y = 17

=> x = 17 - 5 = 12

Therefore, Maria is 12 years old now

Easily we can see the sum, the difference, the product and the quotient of 2 rational number is also rational numbers. 
We have √a + √b + √c = x ( x is a rational number)
=> x - √a = √b + √c => (x² + a - b - c) - 2x√a = 2√bc 
=> (x² + a - b - c)² + 4ax² - 4x(x² + a - b - c)√a = 4bc 
=> √a = [(x² + a - b - c)² + 4ax² - 4bc] / [4x(x² + a - b - c)] => √a is a rational number
Prove the same as above, √b, √c are also rational number

The probability of selecting ace first time is: \(\dfrac{4}{52}\)

The probability of selecting ace second time is: \(\dfrac{4-1}{52-1}=\dfrac{3}{51}\)

In conclusion, the probability of selecting 2 aces is: \(\dfrac{4}{52}\times\dfrac{3}{51}=\dfrac{1}{221}\)

Every day one question

Start from number 1 to number .......

A heptagon has 14 diagonals. 

You mightn't have known:

-A heptagon is a type of polygon, which is a closed plane figure bounded by seven straight edges containing seven angles. When all the sides and angles are equilateral and equiangular, respectively, the heptagon is said to be a regular heptagon. Other types of polygons are triangles, quadrilaterals, pentagons, hexagons and octagons.

- In geometry, a diagonal refers to a side joining nonadjacent vertices in a closed plane figure known as a polygon. The formula for calculating the number of diagonals for any polygon is given as: \(\dfrac{n\left(n-3\right)}{2}\), where n is the number of vertices of the polygon.

Am I clever? Give me a everybody

(SORRY I'M BEING A LITTE BIT SHOWING OFF)

Name the 2 numbers are abc and de

abc - de = 288

=> c - e = 8

=> (c;e) = (8;0),(0;2)

C₁: c = 8; e = 0

=> ab8 - c0 = 288

=> b - c = 8

=> IMPOSSIBLE

C₂: c = 0; e = 2

=> b - (d + 1) = 8

=> b = 7; d = 9

=> a = 3

=> abc + de = 370 + 92 = 462

1x 2 + 3 : 6 x 5 - 4 = 2 + 5/2 - 4 = 1/2 

Their ages could be:

  Possibility                Sum (date)

  1    1    36                     38    

  1    2    18                     21

  1    3    12                     16

   1    4    9                       14

  1     6    6                       13

  2     2    9                       13

  2     3    6                       11

 3      3    4                       10

Of course Ivan knew the date of today, but he needed more info 

=> There are 2 possibilities that have the same sum: (1;6;6) and (2;2;9)

There is a biggest one => (1;6;6) isn't possible because there are two biggest ones

So their ages are 2, 2 and 9

Given that expression as a

We have \(\left(k+1\right)\sqrt{\dfrac{k+1}{k}}>1\) with k = 1, 2, 3, ..., n

Using Cauchy's inequality (bất đẳng thức Cô Si) with k + 1 numbers, we have:

\(\left(k+1\right)\sqrt{\dfrac{k+1}{k}}=\left(k+1\right)\sqrt{\dfrac{1.1.1.....1}{k}.\dfrac{k+1}{k}}< \dfrac{1+1+1+...+1+\dfrac{k+1}{k}}{k+1}=\dfrac{k}{k+1}+\dfrac{1}{k}=1+\dfrac{1}{k\left(k+1\right)}\)

=> \(1< \left(k+1\right)\sqrt{\dfrac{k+1}{k}}< 1+\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)\)

In turn giving k= 1, 2, 3, ..., n then sum them all, we have

\(n< \sqrt{2}+\sqrt{\dfrac{3}{2}}+...+\left(n+1\right)\sqrt{\dfrac{n+1}{n}}< n+1-\dfrac{1}{n}< n+1\)

=> [a] = n