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True [87]
2 years ago
14

5

Mathematics
1 answer:
garik1379 [7]2 years ago
6 0

Given:

The graph of a line.

To find:

The equation of the given line.

Solution:

From the given graph, it is clear that the line passes through the points (0,-2) and (2,0). So, the equation of the line is :

y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)

y-(-2)=\dfrac{0-(-2)}{2-0}(x-0)

y+2=\dfrac{2}{2}(x)

y+2=x

Subtracting 2 from both sides, we get

y+2-2=x-2

y=x-2

Therefore, the correct option is B.

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The distance between the point (2,1,1) and the plane x-2y=5 is​
Eduardwww [97]

The given plane has normal vector

x-2y=5\implies\mathbf n=\langle1,-2,0\rangle

Scaling <em>n</em> by a real number <em>t</em> gives a set of vectors that span an entire line through the origin. Translating this line by adding the vector <2, 1, 1> makes it so that this line passes through the point (2, 1, 1). So this line has equation

\mathbf r(t)=\langle1,-2,0\rangle t+\langle2,1,1\rangle=\langle 2+t, 1-2t, 1\rangle

This line passes through (2, 1, 1) when <em>t</em> = 0, and the line intersects with the plane when

x-2y=5\implies(2+t)-2(1-2t)=5\implies5t=5\implies t=1

which corresponds the point (3, -1, 1) (simply plug <em>t</em> = 1 into the coordinates of \mathbf r(t)).

So the distance between the plane and the point is the distance between the points (2, 1, 1) and (3, -1, 1):

\sqrt{(2-3)^2+(1-(-1))^2+(1-1)^2}=\boxed{\sqrt5}

7 0
3 years ago
At 10 A.M. a plane leaves Boston, Massachusetts, for Seattle, Washington, a distance of 3000 mi. One hour later a plane leaves S
Reil [10]

Answer:

The plane from Seattle will pass the plane from Boston after 4.5 hours of its departure from the Seattle.

Step-by-step explanation:

Plane-1 = Boston, Massachusetts - Seattle, Washington at 10:00 AM

Plane-2= Seattle, Washington - Boston, Massachusetts at 11:00 AM

Speed of plane-1 = 300 mile/hour

Distance covered by plane-1 in 1 hour = 300 mile/h × 1 h = 300 mile

Distance between plane-1 and plane 2 after 1 hour :

= 3000 mile - 300 mile = 2700 mile

Let the distance covered by plane-1 after 1 hour be x in t time where it will pass plane-2.

Let the distance covered by plane-2 be y in t time where it will pass plane-1.

x + y = 2700 ...[1]

Speed of plane-1 = 300 mile/hour

300 mile/h=\frac{x}{t}..[2]

Speed of plane-2 = 300 mile/hour

300 mile/h=\frac{y}{t}..[3]

Putting value of t from [2] in [3];

t=\frac{x}{300 mile/h}

300 mile/h=\frac{y}{\frac{x}{300 mile/h}}

\frac{x}{300 mile/h}=\frac{y}{300 mile/h}

x = y

In [1] put x = y

y + y = 2700 miles

2y = 2700 miles

y = 1,350 mile

For the value of t, put the value of y in [3]:

t=\frac{1,350 mile}{300 mile/h}=4.5 hour

The plane from Seattle will pass the plane from Boston after 4.5 hours of its departure from the Seattle.

5 0
3 years ago
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