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Lina20 [59]
3 years ago
6

Find the coordinates of the midpoint of the segment whose endpoint are H(8,13) K(2,1)​

Mathematics
1 answer:
sesenic [268]3 years ago
7 0

Answer:(

7

,

6

)

Explanation:

We can find the coordinates of the midpoint of a line segment, by finding the mean of the sum of the  

x

coordinates and the mean of the sum of the  

y

coordinates.

x

=

8

+

6

2

=

7

y

=

2

+

10

2

=

6

Coordinate of midpoint:

(

7

,

6

)

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baherus [9]

Step-by-step explanation:

Linear pairs = 180°!!!!!

  1. 1st angle is 2[x]-39 2nd angle is just x
  2. soooo add the x. 3x-39=180
  3. 3x=141 x=47

X is just the second angle!!!!!

lol. i toonk geo last year and i cant believe i still remember all this hopefully i got it right!

8 0
3 years ago
Represent the following expressions as a power of the number a (a≠0): c ((a2)−2)5÷(a4a)3
s2008m [1.1K]

Answer:

1/a²⁷

Step-by-step explanation:

Given: $ \frac{((a^2)^{-2})^5}{a^4.a^3} $

We know that when the bases are same the powers can be added. i.,e.,

                                           (xᵃ)ᵇ = x ⁽ᵃ ⁺ ᵇ⁾

⇒ $ \frac{((a^2)^{-2})^5}{a^4.a^3} \implies \frac{(a^2)^{-10}}{a^7} $

$  \implies \frac{a^{-20}}{a^7} = a^{-20 - 7} = a^{-27} $

Also, $ \frac{x^a}{x^b} = x^{a - b} $

This is nothing but, $ \frac{1}{a^{27}} $.

Note that $ a $ cannot be zero here. The condition is provided in the question as well.

7 0
3 years ago
. Select all the properties of a rectangle.
8090 [49]

Answer:

A, C, D

Step-by-step explanation:

8 0
3 years ago
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If a rink is 3 times its width and the width is 10m what is the area
emmainna [20.7K]
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8 0
3 years ago
Find the midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9)
JulsSmile [24]

The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is \left(\frac{-9}{2}, \frac{-3}{2}\right)

<u>Solution:</u>

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right) is given by:

\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9

Plugging in the values in formula, we get,

\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}

Hence, the midpoint of the segment is \left(\frac{-9}{2}, \frac{-3}{2}\right)

6 0
3 years ago
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