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3 years ago

Which of the following statements is needed in order to prove these triangles are congruent by AAS?

Mathematics

1 answer:

creativ13 [48]3 years ago

5 0

Answer:

$RQ\cong UT$

Step-by-step explanation:

Two triangles are congruent by AAS postulate if two adjacent corresponding angles are congruent and the next adjacent sides to any of the angles is are also congruent. The adjacent sides should not be in between the two congruent angles.

From the triangles RQS and UTV

$\angle R\cong \angle U\\\angle S\cong \angle V$

The adjacent side to $\angle R$ is RQ and for $\angle U$ is UT.

Similarly, the adjacent side to $\angle S$ is QS and for $\angle V$ is TV.

So, the possible sides that could be congruent by AAS postulate are:

$RQ\cong UT$ or $QS\cong TV$

So, the correct option is $RQ\cong UT$

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F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte

Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

(b) Local minimum at (1, 0)

Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

2x² - x -1 = 0

( x - 1) (2x + 1) = 0

x = 1 or x = -½

We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½ = ⁵/₂

ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

x² + 2 = 0

x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

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The function is concave up on (0, ∞).

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3 years ago

Pleaseeee help!!!! I will mark you as brainlinest for correct answer!!!!

pshichka [43]

Yes, your answer is correct.

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3 years ago

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

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 time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

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2 years ago

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telo118 [61]

Answer:

$465

Step-by-step explanation:

100 - 50/3 = 250/3

250/3 ÷ 100 = 5/6

$387.50 ÷ 5/6 = $465

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2 years ago

if the local rate for electrical service is 6cents per kilowatt-hour how much money would be saved in a year by using a dishwash

dezoksy [38]

Given:

The local rate for electrical service is 6 cents per kilowatt-hour.

We need to know how much money would be saved in a year by using a dishwasher 4times per week rather than 6 times per week

From the table:

The cost of using a dishwasher 4 times per week = $29

The cost of using a dishwasher 6 times per week = $44

So, the saving will be = 44 - 29 = 15

So, the answer will be $15

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1 year ago