All categories

3 years ago

3 miles = how many feet

2 answers:

8 0

3 miles=15,840
1 mile= 5,280
So, 1) 5,280
1) 5,280
+1) 5,280
————————-
=15,840
Or 5,280
X. 3
————————-
15,840
Just some examples

LUCKY_DIMON [66]3 years ago

7 0

15,840 feet. best of luck mate

You might be interested in

$4 is what percent of $50 ? show work

hoa [83]

The answer would be 8 although, if you have to show work I can't help you.. I'm trying to help but I haven't learned percentage yet so I am using my percentage calculator.

7 0

3 years ago

What is the answer for this problem 3-3×6+2=?

liberstina [14]

Answer:

-13

Im correct!

6 0

3 years ago

Read 2 more answers

HELP PLEASE FAST!!!!!!!

snow_lady [41]

Answer:

Takeru bought 72 eggs and baked 18 soufflés .

Step-by-step explanation:

Let

e = number of eggs Takeru bought.

s = number of soufflés Takeru bought.

Then we know that Takeru bought 4 times as many eggs as he baked soufflés, therefore:

$e=4s.$

And since for every soufflés Takeru bakes he uses 3 eggs, and after backing he has 18 eggs left, we have:

$e-3s=18$ <em>This says that from $e$ eggs there are 18 eggs left after Tkeru used three eggs for each soufflé.</em>

Now we have two equations:

$(1).e=4s.$

$(2).e-3s=18$

We put the value of e from equation (1) into equation (2) and solve for s and get:

$s=18$

$e=4s=4*18=72.$

Thus

Number of eggs Takeru bought = 72.

Number of soufflés Takeru bought = 18.

3 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

3 years ago

PLSSS HELP IF YOU TURLY KNOW THISS

neonofarm [45]

C ⠀ ⠀ ⠀ ⠀ ⠀

The points between 0 and 1 increase in increments of 1/6th

5 0

2 years ago

Read 2 more answers