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tatuchka [14]
3 years ago
6

Is every rhombus with four right angles is a square. True or False

Mathematics
2 answers:
Step2247 [10]3 years ago
7 0

Answer:

true

Step-by-step explanation:

because the square has 4 congruent sides and 4 right angles

Amanda [17]3 years ago
4 0

Answer:

True.

Step-by-step explanation:

A rhombus has four side of the same length (sometimes) and bisecting diagonals.

A square has what a rhombus has, as well as what a rectangle has.

A rectangle has parallel sides, four right angles, and congruent diagonals.

That means that a square has four right angles, four sides of the same length, and bisecting diagonals.

That's why a square is a rhombus and a rectangle but not its converse.

So, true.

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givi [52]

Rewrite y^4 as (y^2)^2

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Rewrite 49 as 7^2

(y^2)^2 + 14y^2 +7^2

Factor using the perfect square rule.

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6 0
3 years ago
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Kevin ate 4 pies in 132 seconds what is the unit rate ?
solong [7]

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there are 60 teams attend a chess tournament. every team will play with every other team exactly once. Supposed each team has a
Karolina [17]

The probability that no two teams win the same number of games, P ≈ 2.084 × 10⁻⁴⁵³

The reason for arriving at the above probability is as follows:

The given parameters are;

The number of teams in the tournament, n = 60

The chance of a team winning a game = 50% = 0.5

The number of ties = No ties

The required parameter:

The probability that no two teams win the same number of games

Method:

Calculate the number of ways no two teams win the same number of games, and divide the result by the total number of possible outcomes

Solution:

The number of matches played, n = \dbinom {60} {2} = 1,770

The possible outcomes = 2; Winning or losing

The total number of possible outcomes, n_p = 2¹⁷⁷⁰

The number of games won by each team is between 0 and 59

The ways in which no two teams won the same number of games is given by the games won by the teams to be 0, 1, 2,..., 57, 58, 59

Therefore, the number of ways no two teams won the same number of games, the required outcomes, n_k = 59!

Probability = \dfrac{Number \ of \ possible \ outcomes}{Number \ of \ required\ outcomes}

The probability that no two teams win the same number of games is given as follows;

\mathbf{P(No \ two \ teams \ won \ the \ same \ number \ of \ games)} = \dfrac{n_k}{n_p}

Therefore;

P(No \ two \ teams \ won \ the \ same \ number \ of \ games) = \dfrac{59!}{2^{1,770}} \approx \mathbf{2.084 \times 10^{-453}}

The probability that no two teams win the same number of games, P ≈ 2.084 × 10⁻⁴⁵³

Learn more about probability theory here:

brainly.com/question/1539712

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2 years ago
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Vesna [10]
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3 years ago
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The answer is .C (1.035)x
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