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3 years ago

Given: △ABC, AB = CB AK ⊥ BC , MC ⊥ AB Prove: AK = MC

Mathematics

1 answer:

MatroZZZ [7]3 years ago

5 0

By using congruence of two triangles AKC and CMA we proved that

AK = MC

Step-by-step explanation:

Lets revise the cases of congruence

SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ

In Δ ABC

∵ AB = BC

∴ Δ ABC is an isosceles triangle

∴ ∠BAC ≅ m∠BCA ⇒ base angles of isoscelesΔ

∵ AK ⊥ BC

∴ m∠AKC = 90°

∵ MC ⊥ AB

∴ m∠CMA = 90°

∴ ∠AKC ≅ ∠CMA

In the two Δs AKC and CMA

∵ ∠BAC ≅ m∠BCA

∵ ∠AKC ≅ ∠CMA

∵ AC ≅ CA

∴ Δ AKC and Δ CMA are congruent by AAS postulate

∴ AK ≅ CM

By using congruence of two triangles AKC and CMA we proved that

AK = MC

Learn more:

You can learn more about congruence in brainly.com/question/3202836

#LearnwithBrainly

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let X represent the amount of time till the next student will arriv ein the library partking lot at the university. If we know t

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0.606531 = 60.6531% probability that it will take between 2 and 132 minutes for the student to arrive at the library parking lot.

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

$f(x) = \mu e^{-\mu x}$

In which $\mu = \frac{1}{m}$ is the decay parameter.

The probability that x is lower or equal to a is given by:

$P(X \leq x) = \int\limits^a_0 {f(x)} \, dx$

Which has the following solution:

$P(X \leq x) = 1 - e^{-\mu x}$

The probability of finding a value higher than x is:

$P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}$

Mean of 4 minutes

This means that $m = 4, \mu = \frac{1}{4} = 0.25$

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This is:

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2)$

In which

$P(X \leq 132) = 1 - e^{-0.25*132} = 1$

$P(X \leq 2) = 1 - e^{-0.25*2} = 0.393469$

$P(2 \leq X \leq 132) = P(X \leq 132) - P(X \leq 2) = 1 - 0.393469 = 0.606531$

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