All Answers 358090 Answers

The answer is 12.

Looks like it was not properly copied off of a latex using website

try this

https://web2.0calc.com/questions/marvin-the-fly-starts-at-0-0-each-step-marvin-moves

Looks like it was copied from a website, but you are in luck (probably) this problem has probably already been solved

try https://web2.0calc.com/questions/help-asap_57303

I can't see a diagram, but to solve a parrellogram but I suspect that they are right triangles inside of the parrellogram so the side of the rectangle makes up the height of the parrellogram

example:

Is this correct?

this problem was answered here https://web2.0calc.com/questions/counting_95025

I will pin this, if this answer is wrong I will take a shot at it

in the future please provide a diagram

(c)

Now that we know a + b = p/n 2sqrt(3), and that a - b = 4, we can use system of equations to get the individual values of each variable.

Case 1: a + b = 2sqrt(3)

a + b = 2sqrt(3)

a - b = 4

Adding both equations, 2a = 4 + 2sqrt(3), a = 2 + sqrt(3).

Plugging in 2 + sqrt(3) for the top equation, 2 + sqrt(3) - 2sqrt(3) = -b, so b = sqrt(3) - 2.

So for case 1, \(a = 2 + \sqrt{3}\), and \(b = -2 + \sqrt{3}\)

Case 2: a + b = -2sqrt(3)

a + b = -2sqrt(3)

a - b = 4

Adding both equations, 2a = 4 - 2sqrt(3), a = 2 - sqrt(3)

Plugging in 2 - sqrt(3) for the top equation, 2 - sqrt(3) + 2sqrt(3) = -b, so b = -2 - sqrt(3).

So for case 2, \(a = 2 - \sqrt{3}\), and \(b = -2 - \sqrt{3}\)

Hence, we have:

a = 2 + sqrt(3)     =>        b = -2 + sqrt(3)

                           or

a = 2 - sqrt(3)      =>        b = -2 - sqrt(3)

gggg

(b)

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Earlier, we manipulated a^2 + ab + b^2 into (a - b)^2 + 3ab, however you can also manipulate it to (a + b)^2 - ab (do the expansion and see that it works!)

So... a^3 - b^3 = (a - b)[(a + b)^2 - ab] = 52

However, from (a), we learned that a - b = 4, and that ab = -1, substituting in, we have a^3 - b^3 = 4[(a + b)^2 + 1] = 52.

Dividing by 4, we have (a + b)^2 + 1 = 13, a + b = positive/negative square root of (12), which sqrt(12) = 2sqrt(3), so positive negative 2sqrt(3).

Hence, \(a + b = \pm2\sqrt{3}\)

(a)

By difference of cubes, a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Substituting and manipulating, we have a^3 - b^3 = 4[(a - b)^2 + 3ab]

Substituting again, we have a^3 - b^3 = 4(16 + 3ab) = 64 + 12ab = 52

Subtracting and dividing by 12, we have 12ab = 52 - 64 = -12, ab = -1.

Hence, ab = -1

https://web2.0calc.com/questions/counting_62065

Note that that \(\triangle XYD\) has a base of 12. This means that the height of the triangle can be at most 10/3. 

We want to find the shaded region's area over the triangle's area, which is 36. 

The area of the shaded region is the same as the area of the triangle minus the area of the non-shaded region. 

The nonshaded region is a right triangle similar to \(\triangle XYZ\) with a scale factor of 4/9, so its area is \(36 \times ({4 \over 9})^2 = {64 \over 9}\). 

This means that the shaded region has an area of \(36 - {64 \over 9} = {260 \over 9}\), so the probability is just \({{260 \over 9} \over {36}} = \color{brown}\boxed{65 \over 81}\)

this has already been solved by melody and solution completed by arnolde

if you want to view the full solution you can go to https://web2.0calc.com/questions/my-school-s-math-club-has-6-boys-and-8-girls-i-need-to-select-a-team-to-send-to-the-state-math-competition-we-want-6-people-on-the-team-in-how-many-ways_1

Unfortunately Guest's solution is wrong

The answer is 14 diagonals in a heptagon

why?

the formula for determining the number of diagonals on a polygon is n(n - 3)/2

7*4/2 = 14

In the future please try to include diagrams

this problem has already been solved by proyaop

https://web2.0calc.com/questions/geometry_63869

This problem has already been answered :)

https://web2.0calc.com/questions/cphill-help-pls

if that is incorrect I can come back and attempt to recomplete a solution

81/3 = 27

since this is a ratio problem we can say that 27*3 = x

x = 135

This question has already been answered

Hope this gives you the explination you needed :) https://web2.0calc.com/questions/counting-problem_29576

if you provide the graph with an image or a link to the image maybe we can help you

Afraid this is incorrect, but how did you come to that conclusion?

Hint:

Use  probability contour mapping

Let the piece be    x, y-x and  6-y     units ong ... They add up to 6

Let x represents the number of g and y represents the number of r.

As the number of r are two times the number of g so we have:

y=2x..........1

As the price of one g is \$~1 so for x number of g the total price is 1x.

As the price of one r is \$~3.75 so for y number of r the total price is 3.75\times y.

The two equation connstitute the following system of equations:

y=2x

1 x+3.75y=34

Solving the system we get:

The number of g the person performed is 4 and the number of r the person move on is 8.

1 - [(3/12) (2/11) (1/10)] = 1 / 220

2 - [+ 3 (3/12) (2/11) (9/10)] = 27 / 220

3 - [+ (1/8 - (1/8) (3/12) (2/11) (1/10))] = 219 / 1760

4 - [- (1/8) 3 (3/12) (2/11) (9/10)] = - 27 / 1760

5 - The probability is: [1 / 220  +  27 / 220  +  219 / 1760]  -  27 / 1760 = 13 / 55

4-digit permutations==(1, 3, 3, 3),   (2, 2, 3, 3) ==10
5-digit permutations==(1,1,2,3,3), (1, 2, 2, 2, 3), (2, 2, 2, 2, 2)=51

6-digit permutations==(1,1,1,1,3,3), (1,1,1,2,2,3), (1,1,2,2,2,2)=90 

7-digit permutations==(1,1,1,1,1,2,3), (1,1,1,1,2,2,2)==77 

8-digit permutations==(1,1,1,1,1,1,1,3), (1,1,1,1,1,1,2,2)==36 

9-digit permutations==(1,1,1,1,1,1,1,1,2)==9

10-digit permutation ==(1,1,1,1,1,1,1,1,1,1)==1

Grand Total ==10 + 51 + 90 + 77 + 36 + 9 + 1 ==274 permutations.