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serious [3.7K]
3 years ago
13

In how many ways can Anna arrange 2 math books, 3 physics books, and 5 chemistry books on her shelf if all books of the same sub

ject must be adjacent? (The books are distinguishable.)
​
Mathematics
2 answers:
Rina8888 [55]3 years ago
8 0

Answer:

  • 2 maths books can be arranged as (2!=2) ways
  • 3 physics books can be arranged as (3!=6) ways
  • 5 chemistry books can be arranged as (5!=120) ways
  • Books can be arranged in suject-wise in (3!=6) ways

Hence, total ways are (2×6×120×6) ways =8640 ways.

<h2><u>8640</u> is the right answer.</h2>
svetlana [45]3 years ago
7 0

Answer:

  • 8640 ways

Step-by-step explanation:

Arranging the books of each subject:

<u>Math</u>:

  • Combination of 2 books → 2! = 2

<u>Physics</u>:

  • Combination of 3 books → 3! = 6

<u>Chemistry</u>:

  • Combination of 5 books → 5! = 120

We also need to consider the subjects.

There are 3 subjects, kept separately, they will be arranged in 3! = 6 ways

<u>So the total number of combinations is:</u>

  • 2*6*120*6 = 8640
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Find the length of the segment indicated. Round your answer to the nearest tenth if necessary.
guajiro [1.7K]

Answer:

<em>Length of x ⇒ ( About ) 11.5; Option C</em>

Step-by-step explanation:

<em>~ Let us plan this question step-by-step. We know that the line segment with length 13.1 is a radii to the circle, as well as a hypotenuse to a right triangle. Respectively the hypotenuse of another triangle is a hypotenuse as well. By radii ≅, these two part of these two triangles are ≅ ~</em>

1. If these two parts are ≅, the triangle with leg x has a hypotenuse of 13.1 as well ( through ≅ ). This would mean that Pythagorean theorem is applicable for this triangle, as to solve for line segment x.

2. By Pythagorean Theorem ⇒

6.2^2 + x^2 = 13.1^2,

38.44 + x^2 = 171.61,

x^2 = 133.17

<em>Length of x ⇒ ( About ) 11.5</em>

7 0
3 years ago
4.06 Question 9
Jlenok [28]

Answer: B. Graph of 2 lines that intersect at one point. Both lines are solid. One line passes through (-2,2) and (0,3) and is shaded below the line.

y < = 1/2x + 3...(-2,2)                y < = 1/2x + 3....(0,3)

2 < = 1/2(-2) + 3                        3 < = 1/2(0) + 3

2 < = -1 + 3                                3 < = 0 + 3

2 < = 2 (correct)                        3 < = 3 (correct)

The other line passes through points (0,1) and (1,-2) and is shaded above the line.

y > = -3x + 1...(0,1)            y > = -3x + 1...(1,-2)

1 > = -3(0) + 1                   -2 > = -3(1) + 1

1 > = 0 + 1                         -2 > = -3 + 1

1 > = 1 (correct)                 -2 > = -2 (correct)

4 0
2 years ago
Read 2 more answers
Sam is walking across a bridge and accidentally drops an orange into the river basin below
irina1246 [14]

Answer:

We assume that the orange is dropped at t = 0s.

Once the orange is on the air, the only force acting on it is the gravitational force, then the acceleration of the orange is the gravitational acceleration.

A(t) = -32.17 ft/s^2

Where the negative sign is because this acceleration points downwards.

For the velocity equation, we need to integrate over time, we will get:

V(t) = (-32.17 ft/s^2)*t + V0

Where V0 is the initial vertical velocity of the orange, because the orange is accidentally dropped, this initial velocity is equal to zero.

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For the position equation we need to integrate again, this time we get:

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Where P0 is the initial height of the orange, we know that it is 40ft, then the position equation is:

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Now that we know the equation, we can graph it. (you can see the graph below)

Now we also want to find at what time does the orange hit the water.

This happens when:

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We just need to solve that equation for t.

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(1/2)*(32.17 ft/s^2)*t^2 =  40 ft

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t = √(  (40ft)/( (1/2)*(32.17 ft/s^2)) ) = 1.58 s

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tigry1 [53]
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Mariana [72]
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