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2 years ago

Determine if AB¯¯¯¯¯¯¯¯ and CD¯¯¯¯¯¯¯¯ are parallel, perpendicular, or neither. Given: A (−1, 3), B (0, 5), C (2, 1), D (6, −1)

Mathematics

1 answer:

dolphi86 [110]2 years ago

5 0

Answer:

perpendicular

Step-by-step explanation:

To determine if AB and CD are parallel, perpendicular, or neither, we need to get the slope of AB and CD first

Given A (−1, 3), B (0, 5),

Slope Mab = 5-3/0-(-1)

Mab = 2/1

Mab = 2

Slope of AB is 2

Given C (2, 1), D (6, −1)

Slope Mcd = -1-1/6-2

Mcd = -2/4

Mcd = -1/2

Slope of CD is -1/2

Take their product

Mab * Mcd = 2 * -1/2

Mab * Mcd = -1

Since the product of their slope is -1, hence AB and CD are perpendicular

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The rectangle below has an area of c^2-4x-12 square meters and a length of x+2 meters what expression represents the width of th

Korolek [52]

Answer:

Width = $\frac{c^2 - 4x - 12 }{x+2}$

Step-by-step explanation:

All we have to do is to use the method of changing the subject. First we know that length * width is equal to area.

Area = length * width

We have the area and the length but we don't have the width. So we substitute the values we have int the equation and make the width (w) the subject.

(c^2 - 4x - 12) = (x + 2) * w

(c^2 - 4x - 12) = w(x +2)

Divide both sides by (x+2) so w can stand alone.

$\frac{c^2 - 4x - 12}{x+2} = \frac{w(x+2)}{x+2}$

w = $\frac{c^2 - 4x - 12 }{x+2}$

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2 years ago

Benjamin threw a rock straight up from a cliff that was 48 ft above the water. If the height of the rock h, in feet, after t s

fomenos

Answer: It will take 4 seconds for the rock to hit the water.

Step-by-step explanation:

Given : Benjamin threw a rock straight up from a cliff that was 48 ft above the water.

If the height of the rock h, in feet, after t seconds is given by the equation

$h = -16 t^2+ 52 t + 48$

To find : Time taken for the rock to hit the water.

When rock hit the water height becomes zero , i.e. put h = 0 , we get

$-16 t^2+ 52 t + 48 =0$

Divide equation by 2 , we get

$-8t^2+26t+24=0$

The Laue of x for ax²+bx+c =0 is $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

So , $t=\frac{-26\pm\sqrt{(26^{2}-4(-8)(24))}}{2(-8)}$

$t=\dfrac{-26\pm 38}{-16}\\\\ t=-\dfrac{-26+38}{-16}=-0.75\ \ t=\dfrac{-26-38}{-16}=4$

Since t cannot be negative , so t= 4

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3 years ago

100 POINTS!! OMG YOU NEED TO SOLVE THIS!

natita [175]

Answer:

The correct answer is: $\Delta BDE \cong \Delta BFK$ by rule ASA rule of congruence.

Step-by-step explanation:

First let us prove $\Delta BDE \cong \Delta BFK$ by rule ASA (rule of congruence).

Congruent side:

$\overline{BD} \cong \overline{BF}$ (Given)

Congruent angles:

1. By definition of perpendicular,

$\angle{BFK} = 90 \textdegree \ (Since \ \overline{FK} \ is \ perpendicular \ to \ \overline{AB} \ (\overline{FK} \perp \overline{AB} =Given))$

Also,

$\angle{BDE} = 90 \textdegree \ (Given)$

Therefore,

$\angle{BDE} \cong \angle{BFK}$

or you can say,

$\angle{D} \cong \angle{F}$

2. Common angle between $\Delta BDE \ and \ \Delta BFK$ is $\angle{B}$

In a nutshell, in $\Delta BDE$ ,

$\angle{B}$ (Angle)

$\overline{BD}$ (Side)

$\angle{BDE}$ (Angle)

are congruent to the following angle, side and angle of $\Delta BFK$ :

$\angle{B}$ (Angle)

$\overline{BF}$ (Side)

$\angle{BFK}$ (Angle)

Therefore, by ASA rule of congruence, we can say $\Delta BDE \cong \Delta BFK$ .

Since both triangles are congruent, the sides $\overline{ED}$ and $\overline{FK}$ are also congruent.

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Answer:

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Step-by-step explanation:

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