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ratelena [41]
3 years ago
12

A line passes through the point (-2,2) and is parallel to the line of y=1/2(x-5/2) l. What is the equation of the line passing t

hrough point (-2,2) in point slope form?

Mathematics
1 answer:
grin007 [14]3 years ago
3 0
C is your answer. 

It contains the proper points and form. A is not in point-slope form and B has the incorrect slope. 
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ILL GIVE A BRAINLIEST FOR THE ANSWER THAT HELPS ME THE MOSTT.​
Nuetrik [128]
he would be born june 12 1962
4 0
3 years ago
Solve each prportion. (2/y-4)=(y+3/4)
vampirchik [111]

Answer:

y = 5, -4

Step-by-step explanation:

To solve proportions you cross-multiply.

\frac{2}{y-4} =\frac{y+3}{4} \rightarrow 2\times4 = (y-4)\times(y+3) \rightarrow 8=y^2-y-12

Now that we have 8 = y^2 - y - 12, we must make the entire trinomial equal to 0 by subtracting 8 from both sides.

y^2 - y - 20 = 0

Factor the trinomial.

(y - 5)(y + 4) = 0

Make both factors equal to 0 to solve for both values of y.

  • y - 5 = 0 --> y = 5
  • y + 4 = 0 --> y = -4

y = 5, -4

4 0
2 years ago
What are the coordinates of the endpoints of the midsegment for DEF that is parallel to DE
Nutka1998 [239]

Answer:

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right),  \left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

Step-by-step explanation:

Let points D, E and F have coordinates (x_D,y_D),\ (x_E,y_E) and (x_F,y_F).

1. Midpoint M of segment DF has coordinates

\left(\dfrac{x_D+x_F}{2},\dfrac{y_D+y_F}{2}\right).

2. Midpoint N of segment EF has coordinates

\left(\dfrac{x_E+x_F}{2},\dfrac{y_E+y_F}{2}\right).

3. By the triangle midline theorem, midline MN is parallel to the side DE of the triangle DEF, then points M and N are endpoints of the midsegment for DEF that is parallel to DE.

6 0
3 years ago
Q2) The following travel times were measured for vehicles traversing a 2,000 ft. segment of an arterial: Vehicle Travel Time (s)
adelina 88 [10]

Answer:

TMS = 45.78 m/s

SMS = 45.66 m/s

Step-by-step explanation:

Vehicle Travel Time (s)

1 40.5

2 44.2

3 41.7

4 47.3

5 46.5

6 41.9

7 43.0

8 47.0

9 42.6

10 43.3

Time mean speed and space mean speed are parameters used to express the speed of a group of vehicles in traffic flow.

Time mean speed is given as the average speed of all the cars

The speed for each vehicle is calculated by dividing the (2000 ft) distance by the time taken by each vehicle to complete that segment of the journey.

Let vehicle = V

Travel time = time

Speed = s

V | Time | speed (ft/s)

1 | 40.5 | 49.383

2 | 44.2 | 45.249

3 | 41.7 | 47.962

4 | 47.3 | 42.283

5 | 46.5 | 43.011

6 | 41.9 | 47.733

7 | 43.0 | 46.512

8 | 47.0 | 42.553

9 | 42.6 | 46.948

10| 43.3 | 46.189

TMS = (Σsᵢ)/N

Σsᵢ = (49.383+45.249+47.962+42.283+43.011+47.733+46.512+42.553+46.948+46.189)

Σsᵢ = 457.823

N = number of vehicles = 10

TMS = 457.823 ÷ 10 = 45.7823 = 45.782 m/s

b) Space mean speed is given as

SMS = N ÷ [Σ (1/sᵢ)]

V | Time | speed | (1/sᵢ)

1 | 40.5 | 49.383 | 0.02025

2 | 44.2 | 45.249 | 0.02210

3 | 41.7 | 47.962 | 0.02085

4 | 47.3 | 42.283 | 0.02365

5 | 46.5 | 43.011 | 0.02325

6 | 41.9 | 47.733 | 0.02095

7 | 43.0 | 46.512 | 0.02150

8 | 47.0 | 42.553 | 0.02350

9 | 42.6 | 46.948 | 0.02130

10| 43.3 | 46.189 | 0.02165

Σ (1/sᵢ) = (0.02025+0.02210+0.02085+0.02365+0.02325+0.02095+0.02150+0.02350+0.02130+0.02165)

Σ (1/sᵢ) = 0.219

SMS = 10 ÷ [Σ (1/sᵢ)] = 10 ÷ 0.219

SMS = 45.662 m/s

Hope this Helps!!!

4 0
2 years ago
Of the numbers listed below, which are not multiples of 4? 2,4,7,8,12,15,19,24,34
sashaice [31]
7,15,19,34 are not multiples of four. to find this answer out just divide four into them.
6 0
2 years ago
Read 2 more answers
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