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Fantom [35]
3 years ago
5

What is the least common multiple of 12 and 4?

Mathematics
2 answers:
zhenek [66]3 years ago
4 0

Answer: the least common multiple (LCM) of 12 and 4 is 2

Step-by-step explanation:

faltersainse [42]3 years ago
3 0

Answer:

Step-by-step explanation:

2

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Using this data, how many plots would be found on 6 and 1/2?
galben [10]

Answer: 6 full plots I believe

Step-by-step explanation:

4 0
3 years ago
What is the answer for this problem cause i am very confused
EleoNora [17]

41.8, angle y and the measurement angle labeled as just "138.2" are supplementary angles, meaning that when you add them together they equal 180. So all you have to do is subtract 138.2 from 180 and you get your answer which is 41.8.

4 0
3 years ago
Identifying Possible Triangles
Masja [62]

Answer:

3rd triangle can be constructed with dimensions 2,6,7.

Step-by-step explanation:

sum of any two sides > third side.

difference of any two sides < third side

1.

8+5=13 not >14 (no triangle.)

2.

7+8=15 not >15 (no triangle)

3.

2+6=8>7

2+7=9>6

7+6=13>2

7-2=5<6

7-6=1<2

6-2=4<7

so it is a triangle.

4.

6+3=9 not >10 (not a triangle)

5 0
3 years ago
An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.
andrey2020 [161]

Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are E(1)= ^9C_2 places in which the other 2 red balls can be placed.

A red ball drawn third, there are E(3)= ^7C_2 places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are E(5)= ^5C_2 places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are E(7)= ^3C_2 places in which the other 2 red balls can be placed.

The total number of total event is S= ^{10}C_3

The probability that A selects the first red ball is

P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}

P(A \text{wins})=\frac{36+21+10+3}{120}

P(A \text{wins})=\frac{70}{120}

P(A \text{wins})=0.5833

6 0
3 years ago
Suppose a triangle has sides a, b, and c, and let theta be opposite the side of length a. If cos theta &lt; 0, what must be true
katrin2010 [14]
If \cos \theta\ \textless \ 0, then angle \theta is obtuse and triangle with sides a, b, c is obtuse triangle.
In an arbitrary triangle can be only one obtuse angle, and the side which lies opposite to the largest angle is the largest. Then since <span>angle \theta is opposite the side of length a</span>  you can conclude that a>c and a>b.



6 0
3 years ago
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