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Ymorist [56]
3 years ago
15

The coordinates G(7, 3), H(9, 0), I(5, -1) form what type of polygon?

Mathematics
2 answers:
seropon [69]3 years ago
7 0

Answer:

C is the answer

Step-by-step explanation:

DanielleElmas [232]3 years ago
7 0

Answer:

an equilateral triangle

Step-by-step explanation:

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Please answer I'm giving out a thanks, follow, and a brainliest.
Sveta_85 [38]

The mode of a set of numbers is the number that occurs the most.

The numbers in this set are as follows,

26, 17, 9, 12, 18, 21, 9, 14, 18, 23, 15, 7, 20, 18, 17, and 12

26 I

17 II

9 II

12 II

18 III

21 I

14 I

23 I

15 I

7 I

20 I

The number that appears the most is 18!

So the mode of the set of numbers is 18!



3 0
3 years ago
Read 2 more answers
The legend on a map states that 1 cm is 25 km. If you measure 8 cm on the map, how many kilometers would the actual distance be?
Dmitry [639]

Answer:

200km

Step-by-step explanation:

<u>Given </u>

1cm = 25 km 8cm =?

<u>Solve</u>

if 1cm = 25,

then multiply 8 cm by 25.

Which is 200 km

7 0
3 years ago
Pls help I’ve been stuck on this forever!!
andrew-mc [135]

Answer:

triangles always add up to 180 degrees on the interior and because it has equal sides and angles, the triangle is equalateral.

4 0
3 years ago
Help plssss idk what to do!!!!!!!!!
Firdavs [7]

9514 1404 393

Answer:

  y = 1

Step-by-step explanation:

Recognize that 25 and 125 are powers of 5 and rewrite the equation in terms of powers of 5.

The applicable rules of exponents are ...

  (a^b)^c = a^(bc)

  (a^b)/(a^c) = a^(b-c)

  (a^b)(a^c) = a^(b+c)

__

Your equation can be written as ...

  25^4\div5^{5y}=125^y\\\\(5^2)^4\div5^{5y}=(5^3)^y\\\\5^{8-5y}=5^{3y}\\\\8-5y=3y\qquad\text{equate exponents of the same base}

Now this can be solved as an ordinary linear equation.

  8 = 8y . . . . . . add 5y to both sides

  1 = y . . . . . . . divide by 8

The solution is y = 1.

4 0
3 years ago
What is -2 3/4 divided by -1 1/2
Travka [436]

Exact Form:

−11/6

Decimal Form:

−1.83

Mixed Number Form:

−1 5/6

6 0
3 years ago
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