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givi [52]
2 years ago
10

Anyone know this problem???

Mathematics
2 answers:
LekaFEV [45]2 years ago
7 0
Sorry i need points
sammy [17]2 years ago
4 0

Answer:

2

Step-by-step explanation:

\frac{28}{11} = 2.54545454.....

The repeating digits are 54 ← group of 2

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What is the least common multiple of 4, 6, and 12?
Afina-wow [57]

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12

Step-by-step explanation:

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3 years ago
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Outline the process for Inscribing an equilateral triangle in a circle. Perform the construction in GeoGebra, and take a
mezya [45]

Answer:

I don't use Geogebra, but the following procedure should work.

Step-by-step explanation:

Construct a circle A with point B on the circumference.

  1. Use the POINT and SEGMENT TOOLS to create a circle with centre B and radius BA.
  2. Use the POINT tool to mark points D and E where the circles intersect.
  3. Use the SEGMENT tool to draw segments from C to D, C to E, and D to E.

You have just created equilateral ∆CDE inscribed in circle A.

 

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3 years ago
Evaluate 3/5r +5/8swhen r =14 and s =8
Illusion [34]

Step-by-step explanation:

14×5=70

8×8=64

(3/70) + (5/64)

= (3×64)+(5×70)

----‐------------------

70×64

= (192+350)/448

=542/448

= 271/224

5 0
3 years ago
How many different 4-question geometry quizzes can a teacher make if there are 7 different problems to test on?.
stich3 [128]

Answer:

35 quizzes

Step-by-step explanation:

We need to determine how many ways we can choose 4 questions out of 7 to make the quizzes.

We do this by using the formula \displaystyle nCr=\frac{n!}{r!(n-r)!} which describes the number of ways we can choose r objects given n possible choices. So, if n=7 and r=4, then:

_4C_7=\frac{7!}{4!(7-4)!}\\\\_4C_7=\frac{7*6*5*4!}{4!*3!}\\\\_4C_7=\frac{210}{6}\\\\_4C_7=35

Hence, the teacher can make 35 different geometry quizzes.

3 0
2 years ago
Armand ran the 100 yard dash in 17.18 seconds. Arturo's time has an 8 with a value 10 times the value of the 8 in Armand's time.
frosja888 [35]
8====D

He has a value of 64
4 0
3 years ago
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