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SSSSS [86.1K]
2 years ago
12

Find the slope of the line that passes through the points (2,7) and (2,-6). *

Mathematics
1 answer:
Law Incorporation [45]2 years ago
7 0

Answer:

undefined is the answer

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What is the equation of the following line? (0,0) (2,-8)
Ivahew [28]

Use the slope to do


(-8-0)/(2-0)


-8/2 = -4


The slope is -4


y = -4x

6 0
3 years ago
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What is the LCD of 2/7 and 12/5
SashulF [63]

Answer:

10/35 84/35

Step-by-step explanation:

2/7 12/5

7 × 5 = 35

2 × 5 = 10

12 × 7 = 84

10/35 84/35

4 0
3 years ago
A square unit that is 1 unit wide and 1 unit long is a?
MAXImum [283]

Answer:

A square with side length 1 unit, called 'a unit square,' is said to have 'one square unit' of area, and can be used to measure area.

6 0
3 years ago
Which integer is √59 closer to
valentinak56 [21]

It is close to 8. When you look for the square root of 59 you should get about 7.6811457... Round up and thats 8 round down that 7 but it should be 8.

4 0
2 years ago
Read 2 more answers
Points A and B have coordinates (3,1,2) and (1,-5,4) respectively. Point C lies on line AB such that AC:BC=3:2. Find position ve
Sever21 [200]

Answer:

The position vector of point C is <-3 , -17 , 8> or -3i - 17j + 8k

Step-by-step explanation:

* Lets revise how to solve the problem

- If the endpoints of a segment are (x1 , y1 , z1) and (x2 , y2 , z2), and

 point (x , y , z) divides the segment externally at ratio m1 :m2, then

 x=\frac{m_{1}x_{2}-m_{2}x_{1}}{m_{1} -m_{2}},y=\frac{m_{1}y_{2}-m_{2}y_{1}}{m_{1}-m_{2}},z=\frac{m_{1}z_{2}-m_{2}z_{1}}{m_{1}-m_{2}}

* Lets solve the problem

∵ AB is a segment where A = (3 , 1 , 2) and B = (1 , - 5 , 4)

∵ Point C lies on line AB such that AC : BC=3 : 2

∵ From the ratio AC = 3/2 AB

∴ C divides AB externally

- Lets use the rule above to find the coordinates of C

- Let Point A is (x1 , y1 , z1) , point B is (x2 , y2 , z2) and point C is (x , y , z)

 and AC : AB is m1 : m2

∴ x1 = 3 , x2 = 1

∴ y1 = 1 , y2 = -5

∴ z1 = 2 , z2 = 4

∴ m1 = 3 , m2 = 2

- By using the rule above

∴ x=\frac{3(1)-2(3)}{3-2}=\frac{3-6}{1}=\frac{-3}{1}=-3

∴ y=\frac{3(-5)-2(1)}{3-2}=\frac{x=-15-2}{1}=\frac{-17}{1}=-17

∴ z=\frac{3(4)-2(2)}{3-2}=\frac{12-4}{1}=\frac{8}{1}=8

∴ The coordinates fo point c are (-3 , -17 , 8)

* The position vector of point C is <-3 , -17 , 8> or -3i - 17j + 8k

3 0
3 years ago
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