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GREYUIT [131]
3 years ago
9

Is .634 a natural number

Mathematics
1 answer:
kozerog [31]3 years ago
6 0
Fractions and decimals are not natural numbers. Only counting numbers are. ex. 3, 18, and 93. So, no.
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Pls help BRAINLYEST FOR BEST ANSWER YAYA
son4ous [18]

Answer:

A) 21

Step-by-step explanation:

Mean:

(23 + 18 + 24 + 15 + 25)/5

105/5

21

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3 years ago
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For questions 2 and 3 plot the points in the coordinate plane. Then determine whether AB and CD are congruent. Explain your answ
max2010maxim [7]

Answer:

  2. congruent

  3. congruent

Step-by-step explanation:

Segments will be congruent if their lengths are the same. Their lengths will be the same if the sum of squares of the x- and y-differences of their endpoint coordinates are the same.

2. B-A = (-4-(-4), 8-1) = (0, 7)

  D-C = (5-(-2), -5-(-5)) = (7, 0)

The total of 0² and 7² is the same as the total of 7² and 0², so these segments are congruent. (They are both of length 7.)

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3. B-A = (2-4, 6-1) = (-2, 5)

  D-C = (-4-(-2), -3-2) = (-2, -5)

The sum of (-2)² and 5² is the same as the sum of (-2)² and (-5)², so these segments are congruent. (They both are of length √29.)

6 0
4 years ago
There are 6 brown, 5 blue, and 2 orange marbles in a hat. What is the probability of picking two orange marbles in a row without
pashok25 [27]

The total number of marbles = 6 + 5 + 2 = 13 marbles

Orange= 2 marbles

The probability of getting the first marble out of the hat is

2 / 13

The probability of getting the second marble out of the hat is

1 / 12  The total has been reduced by one  and the number of orange marbles has been reduced by one.

2/13 * 1/12 = 2 / 156 = 1/78

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3 years ago
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Step-by-step explanation:

5 0
3 years ago
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Question : In the given figure , ∆ APB and ∆ AQC are equilateral triangles. Prove that PC = BQ.
lorasvet [3.4K]

Answer:

See Below.

Step-by-step explanation:

We are given that ΔAPB and ΔAQC are equilateral triangles.

And we want to prove that PC = BQ.

Since ΔAPB and ΔAQC are equilateral triangles, this means that:

PA\cong AB\cong BP\text{ and } QA\cong AC\cong CQ

Likewise:

\angle P\cong \angle PAB\cong \angle ABP\cong Q\cong \angle QAC\cong\angle ACQ

Since they all measure 60°.

Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:

m\angle PAC=m\angle PAB+m\angle BAC

Likewise:

m\angle QAB=m\angle QAC+m\angle BAC

Since ∠QAC ≅ ∠PAB:

m\angle PAC=m\angle QAC+m\angle BAC

And by substitution:

m\angle PAC=m\angle QAB

Thus:

\angle PAC\cong \angle QAB

Then by SAS Congruence:

\Delta PAC\cong \Delta BAQ

And by CPCTC:

PC\cong BQ

5 0
3 years ago
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