Point c(3.6, -0.4) divides in the ratio 3 : 2. if the coordinates of a are (-6, 5), the coordinates of point b are .

if point d divides in the ratio 4 : 5, the coordinates of point d are .

reset next

Point B is (10 , -4)

Point D is (10/9 , 1)

Step-by-step explanation:

* Lets revise the rule of the point which divides of a line segment in

 a ratio

- If point (x , y) divides the line segment AB, where A is (x1 , y1) and

 B is (x2 , y2) in the ratio m1 : m2

∴ x = [m2(x1) + m1(x2)]/(m1 + m2)

∴ y = [m2(y1) + m1(y2)]/(m1 + m2)

* Now lets solve the problem

- Point C (3.6 , -0.4) divides AB in the ratio 3 : 2, where A is (-6 , 5)

# x = 3.6 , y = -0.4

# A is (x1 , y1) , B is (x2 , y2)

∴ x1 = -6 , y1 = 5

∵ m1 : m2 = 3 : 2

- Substitute these values in the rule

∵ x = [m2(x1) + m1(x2)]/(m1 + m2)

∴ 3.6 = [2(-6) + 3(x2)]/(3 + 2)

∴ 3.6  = [-12 + 3x2]/5 ⇒ multiply both sides by 5

∴ 18 = -12 + 3x2 ⇒ add 12 to both sides

∴ 30 = 3x2 ⇒ divide both sides by 3

∴ 10 = x2

* The x-coordinate of B is 10

∵ y = [m2(y1) + m1(y2)]/(m1 + m2)

∴ -0.4 = [2(5) + 3(y2)]/(3 + 2)

∴ -0.4 = [10 + 3y]/5 ⇒ multiply both sides by 5

∴ -2 = 10 + 3y2 ⇒ subtract 10 from both sides

∴ -12 = 3x2 ⇒ divide both sides by 3

∴ -4 = y2

* The y-coordinate of B is -4

∴ Point B is (10 , -4)

- Point D divides AB in the ratio 4 : 5 where A (-6 , 5) and B (10 , -4)

- To find the coordinates of point D use the same rule above

# D is (x , y)

# A is (x1 , y1) and B is (x2 , y2)

# m1 : m2 is 4 : 5

∵ x1 = -6 and y1 = 5

∵ x2 = 10 and y2 = -4

∵ m1 = 4 and m2 = 5

- Substitute these values in the rule

∵ x = [m2(x1) + m1(x2)]/(m1 + m2)

∴ x = [5(-6) + 4(10)]/(4 + 5) ⇒ multiply the numbers

∴ x = [-30 + 40]/9 ⇒ add

∴ x = [10]/9 ⇒ Divide

∴ x = 10/9

* The x-coordinate of D is 10/9

∵ y = [m2(y1) + m1(y2)]/(m1 + m2)

∴ y = [5(5) + 4(-4)]/(5 + 4) ⇒ multiply the numbers

∴ y = [25 + -16]/9 ⇒ add

∴ y = [9]/9 ⇒ Divide

∴ y = 1

* The y-coordinate of point D is 1

∴ Point D is (10/9 , 1)

.24

step-by-step explanation:

the probability is   .24

the first player had the highest probability of getting a silver marble

the game could be made fair by adding back the marble that was drawn after each draw