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3 years ago

Cual es la ecuación de la recta que pasa por el punto (1, 2) y tiene una inclinación de 71.56505118

Mathematics

1 answer:

kogti [31]3 years ago

3 0

Answer:

La inclinación de una recta esta definida como el ángulo entre el eje x positivo y la recta.

Primero, podemos escribir genéricamente una recta como:

y = p*x + s

donde p es la pendiente de la recta, y s es la ordenada al origen.

Sabemos que para una recta que pasa por los puntos (a, b) y (c, d) podemos escribir la pendiente como:

p = (d - b)/(c - a)

Y también sabemos que si θ es el ángulo de inclinación, entonces:

p = tan(θ)

Entonces en este caso sabemos que:

p = tan(71.56505118°) = -0.828

Entonces esta recta es:

y = -0.828*x + s

y sabemos que pasa por el punto (1, 2), entonces podemos remplazar esos valores en la ecuación para encontrar el valor de s.

2 = -0.828*1 + s

s = 2 + 0.828 = 2.828

La ecuación es:

y = -0.828*x + 2.828

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To find the extreme values of a function f(x,y) on a curve xequalsx(t), yequalsy(t), treat f as a function of the single vari

pychu [463]

Answer:

Absolute maximum is 2

Absolute minimum at -2

Step-by-step explanation:

The given parametric functions are:

$x=2\cos t,y=2\sin t$

By the chain rule:

$f'(t)=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{df}{dt} =\frac{2 \cos t}{-2\sin t} =-\cot(t)$

At fixed points, $f'(t)=0$

$\implies -\cot (t)=0$

This gives $t=\frac{\pi}{2} ,\frac{3\pi}{2}$ on $0\le t\le 2\pi$

This implies that the extreme points are $(2\cos \frac{\pi}{2}, 2\sin \frac{\pi}{2})=(0,2)$ and $(2\cos \frac{3\pi}{2}, 2\sin \frac{3\pi}{2})=(0,-2)$

By eliminating the parameter, we have $x^2+y^2=4$

This is a circle with radius 2, centered at the origin.

Hence (0,2) is an absolute maximum ,at $t=\frac{\pi}{2}$ and (0,-2) is an absolute minimum at $t=\frac{3\pi}{2}$

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3 years ago

Write an equation in slope-intercept form for each line.<br> (4,-1) and (-2,-1)

professor190 [17]

With each line, a slope-intercept relationship (4,-1) and (-2,-1).

Y = 0 ,X = 5.

<h3>What is slope-intercept form?</h3>

The slope-intercept form is just a means of stating a line's equation so that both the slope (steepness) as well as y-intercept (where another line crosses this same vertical y-axis) are obvious. This expression is frequently referred to as y = mx + b.

<h3>According to the given information:</h3>

To begin, are using the slope formula to figure out the average slope between two points here.

Let:

x1 = 4

y1 = -1

x2 = -2

y2 = -1

m = (y2-y1)/(x2-x1)

= (-1 - (-1))/ (4-(-1))

= 0/5

= 0

So the slope is 0.

We must now calculate the y-intercept. This will be accomplished by converting a single of the points and also the slope together into a point-slope linear equation. y2-y1 = m(x2-x1).

Let’s plug in the point (5,0).

So we get y-(-0) = (0/5)(x-(-5)) ⇒ y+0

= (0/5)x + 0 ⇒ y

So,

Y = 0

X = 5

with each line, a slope-intercept relationship (4,-1) and (-2,-1).

Y = 0 ,X = 5

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1 year ago

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Andreas93 [3]

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2 years ago

Help me

dybincka [34]

Answer:

First you have to multiply negative four in parentheses, then add minus two

-4×(-8+3x)=32-12x

2+32-12x=34-12x

34-12x=82

-12x=48

x=-4

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2 years ago

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blondinia [14]

Answer: d-9

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3 years ago

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