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3 years ago

9

Evaluate the arithmetic series described : 2+(-2)+(-6)+(-10)...,510

1 answer:

Alex_Xolod [135]3 years ago

4 0

It goes -4 each time.

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Complete the point slope equation of the line through (-5,4) and (1,6)

jarptica [38.1K]

Answer:

1/3

Step-by-step explanation:

(-5,4)=(x1,y1)

(1,6)=(x2,y2)

Slope(m)=y2-y1

_____

x2-x1

=6-4

____

1+5

=2÷6

=1/3

6 0

3 years ago

Okay so this is all the information given don't I need more?

Kazeer [188]

I think your ok. thats enough info

5 0

3 years ago

WILL MARK BRAINLIST!!!!

Lesechka [4]

Answer:

Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

If 3v/7=6, then v=<br><br> What is the Answer to the problem?

Dmitrij [34]

<u>Let's solve this problem step-by-step</u>:

$\frac{3v}{7} =6\\\frac{3v}{7} *1=6\\\frac{3v}{7} *\frac{7}{7}=6\\ 3v = 6 * 7\\3v=42\\v=14$

<u>Thus v = 14</u>

<u />

<u>Answer: v= 14</u>

<u></u>

Hope that helps!

5 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

3 years ago