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2 years ago

PLEASE HELP!! I beg of yall

Mathematics

1 answer:

Delicious77 [7]2 years ago

5 0

Answer:oop

Step-by-step explanation:

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Julie is digging a square area in her backyard a for a vegtable garden. If the area is 52 square feet, what is the approximate l

AURORKA [14]

Answer:

The correct option will be: (b) 7 feet.

Step-by-step explanation:

Suppose, the length of the garden is $x$ feet.

As the shape of the garden is like a square, then the area of the garden will be: $(length)^2 = x^2$ feet²

Given that, the area is 52 square feet. So, the equation will be.......

$x^2= 52\\ \\ x=\sqrt{52} = 7.211.... \approx 7$

So, the approximate length of the garden will be 7 feet.

4 0

3 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:

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

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

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

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

2 years ago

How do I solve this math problem?

Alina [70]

You would subtract 11 on both sides of the equation.
Then multiply the reciprocal of 1/2 which is 2.
And you should get -16 as your answer

6 0

3 years ago

Read 2 more answers

Pls help!!! evaluate the function for x=m/4

EastWind [94]

Plug in m/4 into the equation:

$8 (\frac{m}{4})^{2} - 4(\frac{m}{4})$

Solve the parentheses in the equation:

$\frac{m}{4}^{2} = \frac{m}{4} \times \frac{m}{4} = \frac{m^{2}}{16}$
$8 \times \frac{m^{2}}{16} = \frac{8m^{2}}{16} = \frac{m^{2}}{2}$

$4 \times \frac{m}{4} = \frac{4m}{4} = m$

Your equation should now look like this:

$\frac{m^{2}}{2} - m$

The answer is D.

6 0

3 years ago

I need to know how to solve this

ahrayia [7]

Answer:

To find angle y you subtract the two angles form 180 degrees because it is a straight line. Then you subtract angle y and 72 from 180 because a triangle adds up to 180 degrees and you get angle x.

Hope this helps, ask if you have any more questions!

8 0

1 year ago