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2 years ago

Find equation of the line that contains the point (4,-2) and is perpendicular to the line y= _2x+8

Mathematics

2 answers:

madreJ [45]2 years ago

6 0

9514 1404 393

Answer:

y = 1/2x -4

Step-by-step explanation:

We presume the given line is ...

y = -2x +8

This is in slope-intercept form, which allows us to determine easily that the slope of this line is -2.

A perpendicular line will have a slope that is the opposite reciprocal of -2:

m = -1/(-2) = 1/2

The y-intercept of the desired line can be found from the point (x, y) = (4, -2) using the equation ...

b = y - mx

b = -2 -(1/2)(4) = -4

Now, we know the slope and y-intercept of the desired perpendicular line through (4, -2), so we can write its equation as ...

y = 1/2x -4

Additional comment

"Slope-intercept form" is ...

y = mx + b . . . . . . where m is the slope and b is the y-intercept

xz_007 [3.2K]2 years ago

3 0

Answer:

y = 1/2x - 4

Step-by-step explanation:

If two lines are perpendicular to each other, they have opposite slopes.

The first line is y = -2x + 8. Its slope is -2. A line perpendicular to this one will have a slope of 1/2.

Plug this value (1/2) into your standard point-slope equation of y = mx + b.

y = 1/2x + b

To find b, we want to plug in a value that we know is on this line: in this case, it is (4, -2). Plug in the x and y values into the x and y of the standard equation.

-2 = 1/2(4) + b

To find b, multiply the slope and the input of x (4)

-2 = 2 + b

Now, subtract 2 from both sides to isolate b.

-4 = b

Plug this into your standard equation.

y = 1/2x - 4

This equation is perpendicular to your given equation (y = -2x + 8) and contains point (4, -2)

Hope this helps!

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A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d

Alenkasestr [34]

Answer:

μ = Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

Distance 1:

- Ff*d₁ = Ef - Ei

⇒ - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ

⇒ - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ (I)

Distance 2:

- Ff*d₂ = Ef - Ei

⇒ - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒ μ*g*d₂ = 0.5*v² (II)

If we apply (I) + (II)

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ (III)

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁ ⇒ vf² = 2*g*Sin θ*d₁ = v²

then (from the equation III) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒ μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒ μ = Sin θ * d₁ / (d₂ - Cos θ*d₁)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK ⇒ - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁ ⇒ d₂ = Sin θ*d₁ / μ

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oksian1 [2.3K]

<h3>Answer: Choice B</h3>

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We can verify this with the following steps

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On the coordinate plane shown, each grid unit represents 10 feet. Polygon QRST has vertices Q(20, 30), R(−10, 30), S(−10, −20),

beks73 [17]

The coordinate plane is missing, so i have attached it.

Answer:

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Area of room = 1200 sq.ft

Step-by-step explanation:

From the cordinate plane attached, we can see the polygon QRST which represents the floor plan of the roo.

Thus;

The length of the sides of the room are;

ST = RQ = 30

QT = RS = 40

Perimeter of a rectangle is the sum of all sides of the rectangle.

Thus, perimeter of the room = (30 × 2) + (40 × 2) = 140 ft

Now, area of a rectangle is the product of two perpendicular sides.

Thus area of room = 40 x 30 = 1200 sq.ft

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