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-BARSIC- [3]
1 year ago
15

What is the midpoint of (-23,-14) and (-18,2)

Mathematics
1 answer:
zloy xaker [14]1 year ago
7 0

Given two coordinates point 1 as (-23,-14) and point 2 as(-18,2).

To find the midpoint we would use the formula below;

\text{midpoint(x,y)}=(\frac{x_1+x_2_{}_{}}{2},\frac{y_1+y_2}{2})

Where x1=-23, x2=-18, y1=-14, y2=2

We would substitute the values into the midpoint formula.

\begin{gathered} \text{midpoint =(}\frac{-23-18}{2},\frac{-14+2}{2}) \\ =(-\frac{41}{2},-6) \end{gathered}

Therefore, the answer is

\text{midpoint}=(-\frac{41}{2},-6)

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Use the simple interest formula to find the unknown quantity. I = prt I = $9 p = $450 r = 4% t =
Andreyy89

Answer:

1/2

just did this on edgenuity:)

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3 years ago
100000000000000000000000000000000000000000+53624785326348756348927563482564328965342875634289573264
Arlecino [84]

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5.3624785e+55

Step-by-step explanation:

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3 years ago
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Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩
NISA [10]

Answer:

\theta = 108.29

Step-by-step explanation:

Given

u =

v =

Required:

Calculate the angle between u and v

The angle \theta is calculated as thus:

cos\theta = \frac{u.v}{|u|.|v|}

For a vector

A =

A = a * b

cos\theta = \frac{u.v}{|u|.|v|} becomes

cos\theta = \frac{.}{|u|.|v|}

cos\theta = \frac{2*6+4*-7}{|u|.|v|}

cos\theta = \frac{12-28}{|u|.|v|}

cos\theta = \frac{-16}{|u|.|v|}

For a vector

A =

|A| = \sqrt{a^2 + b^2}

So;

|u| = \sqrt{2^2 + 6^2}

|u| = \sqrt{4 + 36}

|u| = \sqrt{40}

|v| = \sqrt{4^2+(-7)^2}

|v| = \sqrt{16+49}

|v| = \sqrt{65}

So:

cos\theta = \frac{-16}{|u|.|v|}

cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}

cos\theta = \frac{-16}{\sqrt{2600}}

cos\theta = \frac{-16}{\sqrt{100*26}}

cos\theta = \frac{-16}{10\sqrt{26}}

cos\theta = \frac{-8}{5\sqrt{26}}

Take arccos of both sides

\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})

\theta = cos^{-1}(\frac{-8}{5 * 5.0990})

\theta = cos^{-1}(\frac{-8}{25.495})

\theta = cos^{-1}(-0.31378701706)

\theta = 108.288386087

<em></em>\theta = 108.29<em> (approximated)</em>

4 0
2 years ago
16.415 rounded to the nearest tenth is 16.4.
jeka94

Answer:

false

Step-by-step explanation: the answer is 20

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3 years ago
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Answer:

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