All categories

2 years ago

Complete the following proof

Mathematics

1 answer:

Nadya [2.5K]2 years ago

3 0

Answer:

Step-by-step explanation:

statment/reason

1) HK≅JK / given (S)

2) IK bisects ∠HKJ / given

3) ∠3≅∠4 / angle bisectors form congruent angles (A)

4) HK≅JK / given

5) IK ≅IK / reflexive propriety (S)

6) ΔIHK≅ΔIJK / SAS theorem of congruency

7) ∠1≅∠2 / corsponding parts of congruent Δs are ≅

8) IK bisects ∠HIJ / angle bisectors form congruent angles

You might be interested in

5 to the power of 2 + (10-7) x 2 to the power of 3

Ivahew [28]

Answer:

241

Step-by-step explanation:

= 5^2 + (10 -7)^3 x 2^3 ^ = exponent

= 10 - 7 = 3

= 25 + 3^3 x 2^3

Calculate the exponents

= 25 + 27 x 8

= 27 x 8 = 216

= 25 + 216

= 241

Answer is 241

Hopes this Helps :D

Mark me Brainliest Please : )

6 0

2 years ago

Given p(x) = 2x3 − 8x2 + 13x − 21, ___ is a factor because ___.

sergejj [24]

X - 3 is a factor because p(3) = 0

3 0

3 years ago

Find the value of (27/8)-2/3

Anvisha [2.4K]

The value of (27/8)-2/3 is..

Exact Form: 65/24

Decimal Form: 2.7083...…

Mixed Number Form: 2 17/24

8 0

2 years ago

Read 2 more answers

A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa

UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

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.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so $\mu = \frac{14}{60} = 0.2333$

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081$

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0

3 years ago

Create a number story that involves multiplying the factors twenty-three and fifty. Then, give a written explanation of how you

zloy xaker [14]

Answer:

There are 50 candy bars and your goal is to sell about 23 and one per day how many day would it take bob to sell 23 candybars

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers