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Verizon [17]
3 years ago
13

Please help you guys, Look at the picture attached

Mathematics
1 answer:
Nookie1986 [14]3 years ago
8 0

Answer:

Step-by-step explanation:

if (x,y) is the centroid. Then x=(x1+x2+x3)/3,y=(y1+y2+y3)/3

5.

x=(-2+4+10)/3=4

y=(6+0+6)/3=4

centroid=(4,4)

6.

x=(3+5-2)/3=2

y=(3-1+1)/3=1

centroid=(2,1)

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(Please help ASAP, Will give brainliest)
Lena [83]

Answer:

B) 2

Step-by-step explanation:

a translation is moving an object and not changing how its rotated or its size

4 0
4 years ago
Read 2 more answers
Please help me with question 5
Fantom [35]
Use the left and right triangles. (the two smaller ones -- ACD and BCD)
By similar triangles ACD is similar to BCD
on the left AD/CD is one ratio
On the right CD/DB is the other. These two ratios are equal because they are corresponding parts.

AD / CD = CD / DB
AD = 3
DB = 12
CD = ??
Substitute.
3/CD = CD/12 Cross multiply
36 = CD^2
sqrt(36) = sqrt(CD^2)
CD = 6 <<<<==== Answer.
A <<<<====Answer.
 
7 0
3 years ago
Evaluate using the values given :<br> 3 pm; use m = 4, and p = 3
Goryan [66]

Answer:

36

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Step-by-step explanation:

<u>Step 1: Define</u>

3pm

m = 4

p = 3

<u>Step 2: Evaluate</u>

  1. Substitute:                   3(3)(4)
  2. Multiply:                       9(4)
  3. Multiply:                       36
5 0
3 years ago
The equation of the graphed line is 2x - 3y = 12.
Kazeer [188]

Answer:

x- intercept = 6

Step-by-step explanation:

to find the x- intercept, let y = 0 in the equation and solve for x

2x - 3y = 12

2x - 3(0) = 12

2x - 0 = 12

2x = 12 ( divide both sides by 2 )

x = 6 ← is the x- intercept

7 0
1 year ago
Helllllppp!!!!!!!!!!!!11 marking brainliest!
aalyn [17]

Hello!

Let the unknown number be x.

Subtract 5:

x-5

Double the result: (multiply the result times 2):

2(x-5)

Hope everything is clear.

Let me know if you have any questions!

#KeepLearning

:-)

4 0
2 years ago
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