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slavikrds [6]
2 years ago
15

In triangles DEF and OPQ, ∠D ≅ ∠O, ∠F ≅ ∠Q, and segment DF ≅ segment OQ. Is this information sufficient to prove triangles DEF a

nd OPQ congruent through SAS? Explain your answer.
Mathematics
1 answer:
kenny6666 [7]2 years ago
8 0

In triangles DEF and OPQ, ∠D ≅ ∠O, ∠F ≅ ∠Q, and segment DF ≅ segment OQ; this is not sufficient to prove triangles DEF and OPQ congruent through SAS

<h3>What are congruent triangles?</h3>

Two triangles are said to be congruent if they have the same shape, all their corresponding angles as well as sides must also be congruent to each other.

Two triangles are congruent using the side - angle - side congruency if two sides and an included angle of one triangle is congruent to that of another triangle.

In triangles DEF and OPQ, ∠D ≅ ∠O, ∠F ≅ ∠Q, and segment DF ≅ segment OQ; this is not sufficient to prove triangles DEF and OPQ congruent through SAS

Find out more on congruent triangle at: brainly.com/question/1675117

#SPJ1

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Beth is going to enclose a rectangular area in back of her house. the house wall will form one of the four sides of the fenced i
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The maximum possible area would have a length of 24 feet and width of 12 feet.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depend on other variables while a dependent variable is a variable that depends on other variables.

Let x represent the length and y represent the width, hence:

Since beth has 48 ft fencing and cover 3 sides, hence:

x + 2y = 48  

x = 48 - 2y     (1)

Also:

Area (A) = xy

A = (48 - 2y)y

A = 48y - 2y²

The maximum area is at A' = 0, hence:

A' = 48 - 4y

48 - 4y = 0

y = 12 feet

x = 48 - 2(12) = 24

The maximum possible area would have a length of 24 feet and width of 12 feet.

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A local animal rescue organization receives an average of 0.55 rescue calls per hour. Use the Poisson distribution to find the p
andreyandreev [35.5K]

Answer:

B) 0.894

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

A local animal rescue organization receives an average of 0.55 rescue calls per hour.

This means that \mu = 0.55

Probability that during a randomly selected hour, the organization will receive fewer than two calls.

P(X < 2) = P(X = 0) + P(X = 1)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.55}*(0.55)^{0}}{(0)!} = 0.577

P(X = 1) = \frac{e^{-0.55}*(0.55)^{1}}{(1)!} = 0.317

P(X < 2) = P(X = 0) + P(X = 1) = 0.577 + 0.317 = 0.894

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