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AleksandrR [38]
3 years ago
13

You have seen that the total number of lights in a triangular lighting rig is given by T(n) = n2/2 + n/2, while T(n+1) = n2/2 +

3n/2 + 1. If T(n) times T(n+1) = f(n), what is the reduced form of f(n)?
Mathematics
1 answer:
inn [45]3 years ago
8 0
Based on the scenario, the reduced form of f(n) is : 

f(n) = n^4/4+n^3 + 5n^2/4+n/2

or, it could also be written as

n/4 (n+1) (n^2 + 3n + 2)

hope this helps
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Can someone please help me
lions [1.4K]

Problem 1

<h3>Answer:  Choice B) x = -2</h3>

------------------

Explanation:

Add the equations straight down.

The x terms add to 5x+(-3x) = 2x

The y terms cancel out since 3y+(-3y) = 0y = 0

The right hand sides add to -7+3 = -3

After adding the equations you should get 2x = -4 which solves to x = -2

=====================================================

Problem 2

<h3>Answer:  y = 4</h3>

------------------

Explanation:

Since 5x = 0, this means x = 0 after dividing both sides by 5.

Use this to find y

y = -4x+4

y = -4(0)+4

y = 4

If you want the value of z, then

x = -3y+z+6

0 = -3(4)+z+6

0 = z-6

6 = z

z = 6

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3 years ago
At a football game jasmine bought a soft pretzel for $2.25 and a bottle of water $1.50 she paid with a $5 dollar bill how much c
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3 years ago
Read 2 more answers
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Alik [6]
For apple flavor because they are most=18 so they will get the chance to picked easily.

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3 years ago
What is the area of 10mm 7mm and 16mm
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Step-by-step explanation:

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Answer:

<em>She need </em><em>72 </em><em>index cards.</em>

Step-by-step explanation:

Ms. James has a 6 square foot bulletin board and a 12 square food bulletin board and she wants to cover both boards with index cards without gaps or overlaps.

Each index card has an area of \dfrac{1}{4} square foot.

Dividing the area of each board with the area of the index card will yield the the number of cards she needs to cover up completely.

Let us assume T₁ and T₂ are the number of cards she needs to cover 6 square feet food bulletin board and 12 square feet food bulletin board respectively. So

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