How are the numbers 342 and 324 alike ? How are they different?

Mathematics

2 answers:

vladimir2022 [97]4 years ago

5 0

They are alike as they use the same three digits, are both three figure numbers.

They are different as the numbers are in a different order giving a different number.

Verizon [17]4 years ago

3 0

They are alike because they have the same numbers and they are different because they each have a different amount

You might be interested in

What is -4w + 7 = -10w + 37? Please show me how to get it!

quester [9]

The answer is w=5 how I got it was u add 10 to both sides than minus seven to both sides than u divide and u get w=5

4 0

3 years ago

Given that the acceleration vector is a(t)=(-4cos(2t),-4sin(2t),t)

fiasKO [112]

Answer:

a(t) = (-2sin2, -2cos0, (1/2)

r(t) = (cos2, sin2, 1/6)

Step-by-step explanation:

By definition, given a position vector, r(t),

- The velocity vector v(t) is the derivative of the position vector.

v(t) = r'(t)

- The acceleration vector a(t) is the derivative of the velocity vector.

a(t) = v'(t) = r''(t).

Knowing that integration is the reverse of differentiation, we can obtain the velocity and position vectors from the given acceleration vector.

a(t) = (-4cos(2t), -4sin(2t), t)

Integrating with respect t, we have:

v(t) = (-4/2)sin(2t), (-4/2)(-cos(2t)), (1/2)t²)

at (1, 0, 1), a(t) = (-2sin2, -2cos0, (1/2)(1))

= (-0.0698, -2, 0.5)

Now, we integrate the velocity vector with respect to t to obtain the position vector.

r(t) = (cos(2t), sin(2t), (1/6)t³)

at (1, 1, 1)

r(t) = (cos2, sin2, 1/6)

= (0.999, 0.035, 0.167)

3 0

3 years ago

A multiple choice test consists of four questions, each of which has three choices. Each question has exactly one correct answer

Luden [163]

Answer:

12 ways

Step-by-step explanation

First, we will count the number of ways to get three correct answers:

Notice there is a set of 4 correct answers (one per each one of the four questions in the test). We need to pick three of them. The order does not matter (for example: it is the same to have the arrangement of correct answers: Question 1 , Question 2 and Question 3, than having the arrangement: Question 2, Question 1 and Question 3). Therefore, we are dealing with combinations from a set of 4 elements forming groups of 3 elements:

$\displaystyle C^4_3=\frac{4!}{(4-3)!3!}=\frac{4\cdot3!}{1!3!}=\frac{4}{1!}=4$

Now we need to count the number of ways we can answer incorrectly the remaining fourth question:

We have a total set of 3*4 = 12 incorrect answers (three per each one of the four questions in the test). But notice that once we have picked three correct answers, then we should exclude from the set of incorrect answers we can pick, those of the three questions already answered. So, we are left with just the set of incorrect answers of the remaining question that has not been responded yet. Thus, we just have 3 incorrect answer to pick for the remaining question.

Then using the fundamental counting principle, we multiply the previous results:

Ways to answer correctly three questions * Ways to answer incorrectly the remaining question

$4\cdot3=12$