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Una recta con pendiente -2 ´pasa por el punto P(5, -1). La abscisa del punto Q que está en esa recta es 1. Encuentre la ordenada

oksian1 [2.3K]

Answer:

La ordenada de Q es 7.

Step-by-step explanation:

La ecuación de una recta es:

y = a*x + b

donde a es la pendiente y b es la ordenada al origen.

Para este problema, sabemos que la pendiente es -2

Entonces:

a = -2

y = -2*x + b

También sabemos que esta recta pasa por el punto (5, -1)

Esto significa que cuando evaluamos la ecuación en x = 5, y toma el valor y = -1

Si reemplazamos esos valores en la ecuación, obtenemos:

-1 = -2*5 + b

-1 = -10 + b

-1 + 10 = b

9 = b

Entonces la ecuación de la recta es:

y = -2*x + 9

Sabemos que el punto Q está en la línea, y la abscisa de este punto es 1.

Entonces este punto es Q (1, k)

donde k es la ordenada de este punto.

Porque nuestra línea también pasa por este punto, podemos concluir que cuando x = 1, y = k

Reemplazando eso en la ecuación tenemos:

k = -2*1 + 9 = -2 + 9 = 7

k = 7

La ordenada de Q es 7.

7 0

2 years ago

Select the equation in which the graph of the line has a negative slope and the y intercept equals -5

tatuchka [14]

Answer:

Step-by-step explanation:

Next time, please share the possible answer choices.

Since the slope and y-intercept are given here, we can immediately write out the slope-intercept form of the equation of this line:

y = mx + b becomes y = -1x - 5, where I have arbitrarily chosen the negative slope -1.

7 0

3 years ago

Factor 9x^2+66xy+121y^2

Korvikt [17]

Answer: (3x + 11y)^2

Demonstration:

The polynomial is a perfect square trinomial, because:

1) √ [9x^2] = 3x

2) √121y^2] = 11y

3) 66xy = 2 *(3x)(11y)

Then it is factored as a square binomial, being the factored expression:

[ 3x + 11y]^2

Now you can verify working backwar, i.e expanding the parenthesis.

Remember that the expansion of a square binomial is:

- square of the first term => (3x)^2 = 9x^2
- double product of first term times second term =>2 (3x)(11y) = 66xy
- square of the second term => (11y)^2 = 121y^2

=> [3x + 11y]^2 = 9x^2 + 66xy + 121y^2, which is the original polynomial.

7 0

3 years ago

Find the orthocenter of the triangle with the given vertices: L(2, 2), M(-4, 2), N(-2, 6) a (-2, 4) b (3, -5) c (2, -4) d (-3, 5

Norma-Jean [14]

9514 1404 393

Answer:

(a) (-2, 4)

Step-by-step explanation:

The orthocenter is the point of concurrence of the altitudes of the triangle.

A plot of the triangle vertices shows the triangle to be acute. That means the orthocenter is inside. This eliminates choices B, C, D, which are all points outside the triangle.

__

Conveniently, side LM is horizontal. This means the orthocenter will have the x-coordinate of the vertical line through N, which is -2. Only one answer choice has an x-coordinate of -2:

a (-2, 4)

8 0

2 years ago

Muiltiply 8x^2(x^2+2xy-6y^2) simplify your answer

stepladder [879]

The answer would be 8x^4 + 16x^3y - 48x^2y^2

In order to find this, multiply 8x^2 by each term individually.

8x^2 * x^2 = 8x^4

8x^2 * 2xy = 16 x^3y

8x^2 * -6y^2 = -48x^2y^2

Now you can put them all in a row.

8x^4 + 16x^3y - 48x^2y^2

3 0

3 years ago

20 points answer this randomly! Birthday?