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arlik [135]
4 years ago
15

What are some similarities and differences between the Third Angle Theorem and the Triangle Angle-Sum Theorem?

Mathematics
1 answer:
Alborosie4 years ago
4 0

Answer:

The similarities are;

1) The Third Angle Theorem and the Triangle Angle-Sum Theorem are based on the sum of the angles in a triangle being equal to 180°

2) The Third Angle Theorem and the Triangle Angle-Sum Theorem are used to prove the measure of the third

3) The Third Angle Theorem and the Triangle Angle-Sum Theorem are triangle theorems

The differences are;

1) The Third Angle Theorem is mainly used to prove the similarity of two triangles, while Triangle Angle-Sum Theorem is used to find the measure of the third angle

2) The value of the third angle may not be determined when using the The Third Angle Theorem to prove the similarities between triangles while the value of the third angle is normally determined calculated when the Triangle Angle-Sum Theorem is used to find the third angle given the other two angles in the triangle

Step-by-step explanation:

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Answer:

6

Step-by-step explanation:

24 divided by 4

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5 0
3 years ago
What are m4 and m1.............................................
djverab [1.8K]

Answer:

m<1 = 136

m<4 = 44

Step-by-step explanation:

Exterior angle thm:

<2 + <3 = <4

4x + 2 + 5x - 3 = 8x + 4

9x - 1 = 8x + 4

9x - 8x = 4 + 1

x = 5

m<4:

8x + 4

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8 0
3 years ago
Data collected at Toronto Pearson International Airport suggest that an exponential distribution with mean value 2,725 hours is
Rzqust [24]

Answer:

P(X >2)

And we can calculate this with the complement rule like this:

P(X>2) = 1-P(X

And using the cdf we got:

P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}, x>0

And 0 for other case. Let X the random variable of interest:

X \sim Exp(\lambda=\frac{1}{2.725})

Solution to the problem

We want to calculate this probability:

P(X >2)

And we can calculate this with the complement rule like this:

P(X>2) = 1-P(X

And using the cdf we got:

P(X>2) = 1- [1- e^{-\lambda x}] = e^{-\lambda x} = e^{-\frac{1}{2.725} *2}= 0.480

8 0
3 years ago
I know I just fill in -1 with x but I’m stuck with 2 different answers and this is an exam sooooo I don’t wanna risk it. Can som
son4ous [18]

Answer:

g(-1) = 4

Step-by-step explanation:

Given:

g(x) = \frac{3x^2 - 2x + 7}{2x + 5}

To find g(-1), substitute x = -1

Thus:

g(-1) = \frac{3(-1)^2 - 2(-1) + 7}{2(-1) + 5}

g(-1) = \frac{3 + 2 + 7}{-2 + 5}

= \frac{3 + 2 + 7}{-2 + 5}

= \frac{12}{3}

g(-1) = 4

7 0
3 years ago
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