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3 years ago

Simplify each expression. use positive exponents. 1. m^3n^–6p^0

Mathematics

1 answer:

loris [4]3 years ago

8 0

$m^3n^{-6}p^0= \frac{m^3*1}{n^6} = \frac{m^3}{n^6}$

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You and your friends going to the movies the cost of admission is $9.50 per person create a table showing the relationship betwe

kvv77 [185]

No. of People Cost

1 9.50

2 19.00

3 28.50

4 38.00

etc.

If we divide the cost in each row by the number of people in that row, we always have the proportion (ratio) of 9.50 / 1

4 0

3 years ago

Given that rectangle MNOP~ rectangle STUV~, what is the length of ___<br> TU

IrinaK [193]

A parallelogram in which adjacent sides are perpendicular to each other is called a rectangle. The value of x or the length of TU is 22.5.

<h3>What is a rectangle?</h3>

That parallelogram in which adjacent sides are perpendicular to each other is called a rectangle. A rectangle is always a parallelogram and a quadrilateral but the reverse statement may or may not be true.

Since all the rectangles are similar, therefore, the corresponding sides of the rectangle will be in ratio. Therefore,

$\dfrac{\text{Length of STUV}}{\text{Length of MNOP}} = \dfrac{9}{6}$

Similarly, the breadth will be,

$\dfrac{\text{Width of STUV}}{\text{Width of MNOP}} = \dfrac{\text{Length of STUV}}{\text{Length of MNOP}}\\\\\\\dfrac{\text{Width of STUV}}{\text{Width of MNOP}} = \dfrac{9}{6}\\\\\\\dfrac{x}{15}=\dfrac96\\\\\\x = \dfrac{9 \times 15}{6} = 22.5$

Hence, the value of x or the length of TU is 22.5.

Learn more about Rectangle:

brainly.com/question/15019502

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5 0

2 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$