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3 years ago

Write an equation of the line that passes through (-4, -1) and is perpendicular to the line y= 4/3x+6.

Mathematics

1 answer:

scoundrel [369]3 years ago

3 0

Answer:

y = (-3/4)x - 4

Step-by-step explanation:

The slopes of perpendicular lines are opposite reciprocals of each other. In other words, if the slope of one line is a/b, then the slope of the line perpendicular to it would be -b/a.

Here, the given slope is 4/3, so the slope of the perpendicular line is -3/4.

We are given a point (-4, -1) and we know the slope, so we can find the point-slope form of the line. Point-slope form is written as y - y1 = m(x - x1), where (x1,y1) is the point and m is the slope. Here, x1 = -4 and y1 = -1 and m = -3/4. So:

y - y1 = m(x - x1)

y - (-1) = (-3/4) * (x - (-4)) = (-3/4) * (x + 4)

y + 1 = (-3/4)x + (-3/4) * 4 = (-3/4)x - 3

y = (-3/4)x - 3 - 1

y = (-3/4)x - 4

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On a hot day, Ava poured 1/4 of a bucket of water into a plastic wading pool.a few minutes later she added another 1/3 of a buck

faltersainse [42]

Answer:

7/12

Step-by-step explanation:

4 x 3 = 12 (LCD)

3/12 + 4/12 = 7/12

No simplifying done,

7/12 is the answer.

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5 0

2 years ago

3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k

Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

$\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt$

Evaluate the integral to solve for y :

$\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt$

$\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x$

$\displaystyle y = x^3+2x^2+kx - 2$

Use the other known value, f(2) = 18, to solve for k :

$18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}$

Then the curve C has equation

$\boxed{y = x^3 + 2x^2 + 2x - 2}$

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

$\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2$

The slope of the given tangent line $y=x-2$ is 1. Solve for a :

$3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1$

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

$x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1$

So, the point of contact between the tangent line and C is (-1, -3).

7 0

2 years ago

Dean and his friends went to Aqua World Water Park, where they floated down the Lazy River for 1 2 of an hour. This was 1 4 of t

Kisachek [45]

1/2 of an hour is 30 minutes
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3 0

2 years ago

In the figure, x ∥ y and a is a transversal that crosses the parallel lines.

Dominik [7]

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In short, Your Answer would be Option C

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3 years ago

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almond37 [142]

Answer:

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2 years ago