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2 years ago

6

What is the equation of the following line? Be sure to scroll down first to see

1 answer:

8_murik_8 [283]2 years ago

5 0

The required equation of the line is 5y=x.

We are given to find the equation of the line passing through the points (0, 0) and (15, 3).

We know that

<h3>What is the slope point form of the line?</h3>

The equation of a line passing through two points (a, b) and (c, d) is given by

$y-b=\frac{d-b}{c-a}(x-a)$

Therefore, the equation of the line passing through the points (0, 0) and (15, 3) is given by,

Thus, the required equation of the line is

$y-0=\frac{3-0}{15-0}(x-0)\\y=\frac{3}{15}x\\ y=\frac{1}{5}x\\ 5y=x$

Thus the required equation of the line is 5y=x.

To learn more about the equation of the line visit:

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Solution for: <br> -x + 3y -2z = 19<br> 2x + y - z = 5<br> -3x - y + 2z = -7

STatiana [176]

Answer:

x = -1 , y = 4 , z = -3

Step-by-step explanation:

Solve the following system:

{-x + 3 y - 2 z = 19 | (equation 1)

2 x + y - z = 5 | (equation 2)

-3 x - y + 2 z = -7 | (equation 3)

Swap equation 1 with equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

2 x + y - z = 5 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Add 2/3 × (equation 1) to equation 2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y/3 + z/3 = 1/3 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Multiply equation 2 by 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

-x + 3 y - 2 z = 19 | (equation 3)

Subtract 1/3 × (equation 1) from equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

0 x+(10 y)/3 - (8 z)/3 = 64/3 | (equation 3)

Multiply equation 3 by 3/2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y + z = 1 | (equation 2)

0 x+5 y - 4 z = 32 | (equation 3)

Swap equation 2 with equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+y + z = 1 | (equation 3)

Subtract 1/5 × (equation 2) from equation 3:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+0 y+(9 z)/5 = (-27)/5 | (equation 3)

Multiply equation 3 by 5/9:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y - 4 z = 32 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Add 4 × (equation 3) to equation 2:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+5 y+0 z = 20 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Divide equation 2 by 5:

{-(3 x) - y + 2 z = -7 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Add equation 2 to equation 1:

{-(3 x) + 0 y+2 z = -3 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(3 x)+0 y+0 z = 3 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Divide equation 1 by -3:

{x+0 y+0 z = -1 | (equation 1)

0 x+y+0 z = 4 | (equation 2)

0 x+0 y+z = -3 | (equation 3)

Collect results:

Answer: {x = -1 , y = 4 , z = -3

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Answer:

The swimmer must move in the direction 338.2°

Step-by-step explanation:

Since the river flows due south at 1.2 miles per hour and the swimmer swims 3 miles per hour due east of the river, since their directions are perpendicular, the resultant speed is the hypotenuse of the triangle formed by the two perpendicular directions.

The direction of the resultant speed is thus θ = tan⁻¹( vertical component/horizontal component) where vertical component = 1.2 mph due south = -1.2 mph and horizontal component = 3 mph due east = + 3 mph

So, θ = tan⁻¹(-1.2 mph/+ 3mph)

θ = tan⁻¹(-1.2/3)

θ = tan⁻¹(-0.4)

θ = -21.8°

θ = -21.8° + 360° (since we are in the fourth quadrant- between east and south)

θ = 338.2°

So, <u>the swimmer must move in the direction 338.2°</u>.

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3 years ago