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!100 POINTS PLEASE ANSWER QUICK NO LINKS!

Mkey [24]

Answer:

4 units

Step-by-step explanation:

Calculate the distance d using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = (- 2, - 4 ) and (x₂, y₂ ) = (- 2, - 8 )

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

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

= $\sqrt{0^2+(-4)^2}$

= $\sqrt{0+16}$

= $\sqrt{16}$

= 4

4 0

3 years ago

Read 2 more answers

Which method is the most effective method to use to solve x^2+4x+3=0

lukranit [14]

Answer:

Completing the square.

Step-by-step explanation:

since there is no "a" term and the bx term is even, completing the square is your best bet.

6 0

3 years ago

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2(k+ 4) = 3(k-2)<br> HELP PLEASE

dybincka [34]

Answer:

k = 14

Step-by-step explanation:

k = 14 because if you isolatethe variable by dividing each side by factor that dosent contain the variable

6 0

3 years ago

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On a coordinate plane, parallelogram H I J K is shown. Point H is at (negative 2, 2), point I is at (4, 3), point J is at (4, ne

Mrrafil [7]

<h3>Answer: Choice C. (1,0)</h3>

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Explanation:

Point H is at (-2,2) and J is at (4,-2)

Focusing on the x coordinates, the midpoint is (x1+x2)/2 = (-2+4)/2 = 1.

The y coordinates have a midpoint of (y1+y2)/2 = (2+(-2))/2 = 0.

The midpoint is at (1,0). That's why the answer is choice C.

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This next section is optional.

Points I and K are located at (4,3) and (-2,-3) respectively

Averaging the x coordinates gets us (x1+x2)/2 = (4+(-2))/2 = 1

Doing the same for the y coordinates gets us (y1+y2)/2 = (3+(-3))/2 = 0

We end up with the same midpoint (1,0)

The midpoint being the same for both diagonals proves that each diagonal is bisected, ie cut in half.

3 0

3 years ago

Assignment The operation teeter is defined on the set T= { 2, 3, 5, 7 } by x teeter y=[x+y+xy) mod.8 a - Construct mod.8 table f

rosijanka [135]

Answer:

Case A) tau_net = -243.36 N m, case B) tau_net = 783.36 N / m, tau_net = -63.36 N m, case C) tau _net = - 963.36 N m,

Explanation:

For this exercise we use Newton's relation for rotation

Σ τ = I α

In this exercise the mass of the child is m = 28.8, assuming x = 1.5 m, the force applied by the man is F = 180N

we will assume that the counterclockwise turns are positive.

case a

tau_net = m g x - F x2

tau_nett = -28.8 9.8 1.5 + 180 1

tau_net = -243.36 N m

in this case the man's force is downward and the system rotates clockwise

case b

2 force clockwise, the direction of

the force is up

tau_nett = -28.8 9.8 1.5 - 180 2

tau_net = 783.36 N / m

in case the force is applied upwards

3) counterclockwise

tau_nett = -28.8 9.8 1.5 + 180 2

tau_net = -63.36 N m

system rotates clockwise

case c

2 schedule

tau_nett = -28.8 9.8 1.5 - 180 3

tau _net = - 963.36 N m

3 counterclockwise

tau_nett = -28.8 9.8 1.5 + 180 3

tau_net = 116.64 Nm

the sitam rotated counterclockwise

4 0

1 year ago