Find the point, Q, along the directed line segment AB that divides AB into the ratio 2:3. The 2:3 ratio means that the line
should be broken up in to 5 equal sections (2 + 3 = 5). This
means that each of the 5 sections can be represented by the
expression AB/5. Therefore, the point that divides AB into the
ratio 2:3 is the distance (AB/5)(2) from A.

Point Q is at a distance of 4.7 units from A.

Step-by-step explanation:

From the graph, AC = 10 units and BC = 6 units. Applying the Pythagoras theorem,

= +

      = +

      = 100 + 36

     = 136

AB =

AB = 11.6619

AB = 11.66

     ≅ 11.7 units

But point Q divides AB into ratio 2:3. Therefore:

AQ = × AB

     =   × 11.66

     = 4.664

AQ = 4.664

AQ ≅ 4.7 units

QB = × AB

     =   × 11.66

     = 6.996

QB  ≅ 7.0 units

So that point Q is at a distance of 4.7 units from A.

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .