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4 years ago

Find the equation of the straight line joining the points (-5,2) and (3,-4)

Mathematics

2 answers:

Murljashka [212]4 years ago

7 0

Answer:

y= -0.75x+5.75

Step-by-step explanation:

so, if y=mx+c

the gradient is m, which is equal to (y1-y2)/(x1-x2), (2--4)/(-5-3) = 6/-8=-0.75

so y= -0.75x+c

substitute y and x values

2(y)= -0.75(m)*-5(x)+c

2=-0.75*5+c

2= -3.75+c

5.75=c

so in the end, y= -0.75x+5.75

Anastaziya [24]4 years ago

6 0

Answer:

3x+4y+7=0.... equations (1)

which is required equations

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A man is standing at a radar base and observes an unidentified plane at an altitude 6000m flying towards the radar base at an an

liubo4ka [24]

Answer:

The speed in of the plane is 115.47 m/sec

Step-by-step explanation:

Given:

Height at which the plane is flying = 6000 m

Angle of elevation at the radar base = 30 Degrees

Angle of elevation at the radar base after one minute = 60 Degrees

To Find:

The Speed of the plane in meter per second = ?

Solution:

Let us use the tangent of the angle to find the distance (d) to a point directly below plane:

when the angle is 30 degrees

$tan(30) = \frac{6000}{d1}$

$d1 = \frac{6000}{tan(30)}$

$d1 = \frac{6000}{0.577}$

d1 = 10392.3 meters

when the angle is 60 degrees

$tan(60) = \frac{6000}{d2}$

$d2 = \frac{6000}{tan(60)}$

$d2 = \frac{6000}{1.732}\\$

d2 = 3464.1 meters

distance travelled by aircraft in 1 min is

=>d1 - d2

=>0392.3 - 3464.1

= 6928.2 m/min

Now converting to m/sec

=> $\frac{6928.2}{60}$

=>115.47 m/sec

4 0

3 years ago

Please solve this problem 5y+30=5()

nadya68 [22]

First you would subtract 30 from both sides
they divide each side by 5 to get -7
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3 years ago

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jeyben [28]

Sorry but I don’t understand your language.

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Explanation:

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3 years ago

Find the equation of the straight line joining the points (-5,2) and (3,-4)​

Find the equation of the straight line joining the points (-5,2) and (3,-4)