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3 years ago

How many different ways can 50 players in a marching band be arranged in rectangle arrangement?

2 answers:

6 0

Since the band has to be arranged in a rectangle arrangement, the way in which the players are going to be organized are going to be factors of 50.
1st way: 1 by 50
2nd way: 2 by 25
3rd way: 5 by 10

Hope this helped!!

PilotLPTM [1.2K]3 years ago

4 0

1*50
2*25
5*50
You can basically do any number that can go in 50

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Please help! i think the answer is 12 but any verification would be greatly appreciated!

ANEK [815]

Answer:

13.9 or 14

Step-by-step explanation:

a^2 + b^2 = c^2

5^2 + 13^2 = c^2

25 + 169 = 197

square root 197 to get c

c = 13.9 (14 rounded up)

Helpful thing to note is that the hyptoenuse will always be longer than your "long side" of the triangle

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3 years ago

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Find the surface area of the rectangular prism. 8 m 4 m 1 m

Dmitry_Shevchenko [17]

Answer:

A=2(wl+hl+hw)

Step-by-step explanation:

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2 years ago

`PLSS PLSSS HELPPP I WILL GIVE BRAINLIESTTTT!!!!!!!

lidiya [134]

Answer:

ASA i think

Step-by-step explanation:

because if you start anywhere on the triangle you start at an angle

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3 years ago

a web designer charges a $200 fee plus $50 per hour to build a website. which equation represents the total cost, y, to a custom

dezoksy [38]

Y = 50x + 200 <=== ur equation

6 0

3 years ago

Read 2 more answers

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

Using partial fraction decomposition to find the definite integral of:

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx$

By using the long division method; we have:

$x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }$

$- 2x^3 -16x^2-40x$

$x+ 20$

So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

Now; the integral is:

$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

$\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}$

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0

2 years ago