What is the slope of a line perpendicular to the line whose equation is 2x−10y=100. Fully reduce your answer.

hjlf

Solve each of the equations independently, then determine if the are continuous or discontinuous.
15≥-3x [start here]
-5≤x [divide both sides by (-3). *Dividing by a negative number means the direction of the sign changes!]
x≥-5 [just turned around for analysis]

Next equation:
2/3x≥-2 [start here]
x≥-2(3/2) [multiply both sides of the equation by the reciprocal, 3/2)
x≥-3

So, (according to the first equation) all values of x must be greater than, or equal to -5.

(According to the second equation) all values of x must be greater than, or equal to -3.
So, when graphed on a number line, both equations graph in the same direction, so they are continuous.

4 0

2 years ago

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Hi uwu pls help me?? i really need it :'( dont make me cry

densk [106]

Answer:

4z-(3z) z

Step-by-step explanation:

I hope that helps

8 0

2 years ago

The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?

Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$A(-6, 2)) = (x_1, y_1)$

$B(2, 6) = (x_2, y_2)$

$AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}$

$AB = \sqrt{(8)^2 + (4)^2}$

$AB = \sqrt{64 + 16} = \sqrt{80}$

$AB = 8.9$ (nearest tenth)

Distance between B(2, 6) and C(7, 1):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(2, 6) = (x_1, y_1)$

$C(7, 1) = (x_2, y_2)$

$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

$BC = \sqrt{(5)^2 + (-5)^2}$

$BC = \sqrt{25 + 25} = \sqrt{50}$

$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

$CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}$

$CD = \sqrt{(-4)^2 + (-6)^2}$

$CD = \sqrt{16 + 36} = \sqrt{52}$

$CD = 7.2$ (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

$DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

$DA = \sqrt{(-9)^2 + (7)^2}$

$DA = \sqrt{81 + 49} = \sqrt{130}$

$DA = 11.4$ (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

8 0

2 years ago

Solve for x,y, and z.

Ann [662]

The given triangle is isosceles, so the two remaining angles in the triangle both have measure xº. The interior angles of any triangle sum to 180º, so that

58º + xº + xº = 180º

58 + 2x = 180

2x = 122

x = 61

Angles y and z are supplementary to angle x, so that

xº + yº = 180º

and

xº + zº = 180º

and consequently, y = z. In particular, we get

y = 180 - 61

y = 119

and so

z = 119

6 0

2 years ago

Select all the numbers that are common multiples of 4 and 6.

Verdich [7]

Answer:

12, 24, and 60.

Step-by-step explanation:

It says multiples. Not factors, so it can't be 1 or 2. Neither of the numbers can make the number 10. 4 can make 40, but 6 can't.

8 0

2 years ago

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