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3 years ago

8

Enter an equation in point-slope form for the line. Slope is -4 and (1,5) is on the line.

1 answer:

Yuki888 [10]3 years ago

4 0

Answer:

ddddddddddddaaaaaaaaa

Step-by-step explanation:

dddddddddasdasdasdsadasdasdsadsa

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What is the length of the line? <br><br>1. √11<br><br>2.√61<br><br>3. 8<br><br>d. 11

scoundrel [369]

Answer:

3.8

Step-by-step explanation:

only one that made sense

5 0

3 years ago

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if abc def and mno pqr then abc pqr by the transitive property true or flase but one trinagle is left out DEF

tiny-mole [99]

Answer:

true

Step-by-step explanation:

4 0

3 years ago

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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

3 years ago

Solve for B -(12 x 2) x (-6) = -12 x (2 x b)<br> A. -6<br> B. 6<br> C. -12<br> D. 12

myrzilka [38]

<u><em>Answer:</em></u>

<u><em>Option A</em></u>

<u><em>Step-by-step explanation:</em></u>

<u><em>-(12 x 2) x (-6) = -12 x (2 x b)</em></u>

<u><em>=> -24 x -6 = -12 x 2b</em></u>

<u><em>=> 144 = -24b</em></u>

<u><em>=> b = -6</em></u>

<u><em>Hoped this helped.</em></u>

5 0

3 years ago

Solve the equation for x<br> 2 (2.0 – 3) = 10

PolarNik [594]

Step-by-step explanation:

this is your answer...........

7 0

3 years ago