Plzzz help <br> HWK#13B<br> Factors

What is the area of the polygon?

astra-53 [7]

Honestly like just count the squares inside

3 0

3 years ago

Read 2 more answers

2. Given a directed line segment with endpoints A(3, 2) and B(6, 11), what is the point that

KATRIN_1 [288]

Answer:

The point of division is (5 , 8)

Step-by-step explanation:

* Lets explain how to solve the problem

- If point (x , y) divides the line whose endpoints are $(x_{1},y_{1})$

and $(x_{2},y_{2})$ at ratio $m_{1}:m_{2}$ , then

$x=\frac{x_{1}m_{2}+x_{2}m_{1}}{m_{1}+m_{2}}$ and

$y=\frac{y_{1}m_{2}+y_{2}m_{1}}{m_{1}+m_{2}}$

* Lets solve the problem

- The directed line segment with endpoints A (3 , 2) and B (6 , 11)

- There is a point divides AB two-thirds from A to B

∵ The coordinates of the endpoints of the directed line segments

are A = (3 , 2) and B = (6 , 11)

∴ $(x_{1},y_{1})$ is (3 , 2)

∴ $(x_{2},y_{2})$ is (6 , 11)

∵ Point (x , y) divides AB two-thirds from A to B

- That means the distance from A to the point (x , y) is 2/3 from

the distance of the line AB, and the distance from the point (x , y)

to point B is 1/3 from the distance of the line AB

∴ $m_{1}:m_{2}$ = 2 : 1

∵ $x=\frac{(3)(1)+(6)(2)}{2+1}=\frac{3+12}{3}=\frac{15}{3}=5$

∴ The x-coordinate of the point of division is 5

∵ $y=\frac{(2)(1)+(11)(2)}{2+1}=\frac{2+22}{3}=\frac{24}{3}=8$

∴ The y-coordinate of the point of division is 8

∴ The point of division is (5 , 8)

6 0

3 years ago

In a fraternity with 40 members , 18 take mathematics, 5 take both mathematics and physics, and 8 take neither mathematics nor p

garri49 [273]

Answer: 14 members take physics but not mathematics

Step-by-step explanation:

Please see the attached for a diagrammatic representation of this situation.

Since there are 40 members but 8 did not offer any of mathematics or physics, then:-

40 - 8 = 32

So, 32 members of the fraternity offer maths, physics or both.

5 members offer both mathematics and physics .

Since 5 members offer both mathematics and physics, the number of members that offer maths alone = 18 - 5

Let us use "P" to represent the number of members that offer physics. Therefore, the number of members that offer only physics = (P - 5)

Now, if we add up the members that offered maths only, the members that offered both mathematics and physics and the members that offered only physics, it will be equal to the 32 members that offered at least a subject.

Therefore, (18 - 5) + 5 + (P - 5) = 32

18 - 5 + 5 + P - 5 = 32

18 + P - 5 = 32

P + 13 = 32

P = 32 - 13

P = 19

19 members offered physics out of the 32 members that offered maths, physics or both.

To find out the number of people or members that offered physics only, we subtract the number of members that offered the two subjects (5) from a total of 19 members that offered Physics

19 - 5 = 14

Therefore, 14 members of the fraternity offer only physics