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3 years ago

Jim ran 8/9 mile on Monday and 6/9 of a mile on Wednesday. How much farther did he run on Monday than on Wednesday

Mathematics

1 answer:

zubka84 [21]3 years ago

4 0

(8/9)-(6/9)=2/9 of a mile farther

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Athena uses her credit card to buy some clothes for $738.14. She can pay off up to $265 per month. The card has an annual rate o

s344n2d4d5 [400]

You can use the following online calculator to answer your question. http://www.arachnoid.com/lutusp/finance.html you would make the folowing entries: Present Value = 738.14 Future Value = 0 Number of Payments is left blank because you want to solve for that. Payment Amount = -265.00 Interest Rate per period = 17.9% per year divided by 12 = 1.491666667% per month. Payment At is at End. click on the NP button and it will tell you the number of months required for the loan to be fully paid off. the calculator will show you that it will take 2.87 months for you to fully pay off the loan of $738.14 when you are paying it off at $265.00 at the end of each momth and the interest rate if 1.491666667% per month. now that you know how long it takes for you to pay it off, you can determine what the interest amount is that you paid. multiply your payment of 265 per month by 2.87 months to get a total payment of $ 760.55 your loan was for $738.14. the difference is the amount of interest you paid. that will be $760.55 - $738.14 =<span> <span>22.41</span></span>$\ you paid $22.41 in interest.

3 0

3 years ago

What is the reciprocal of 2 3/4

lora16 [44]

0.36363636363 is the answer

4 0

3 years ago

Can anyone help me with this? thank you

wel

"MIDDLE OF THE NIGHT"

I summoned you, please come to me

Don't bury thoughts that you really want

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Within me lies what you really want

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'Cause you know this

'Cause you know this sound

In the middle of the night

Just call my name

I'm yours to tame

In the middle of the night

I'm wide awake

I crave your taste all night long

'Til morning comes

I'm getting what is mine

You gon' get yours, oh no, ooh

In the middle of the night

In the middle of the night, oh

These burning flames, these crashing waves

Wash over me like a hurricane

I captivate, you're hypnotized

Feel powerful, but it's me again

Come, lay me down

'Cause I know this

'Cause I know this sound

In the middle of the night

Just call my name

I'm yours to tame

In the middle of the night

I'm wide awake

I crave your taste all night long

'Til morning comes

I'm getting what is mine

You gon' get yours, oh no, ooh

In the middle of the night

In the middle of the night, oh

Just call on me, ah

Just call my name

Like you mean it, ah

In the middle of the night

Just call my name

I'm yours to tame

In the middle of the night

I'm wide awake

I crave your taste all night long

'Til morning comes

I'm getting what is mine

You gon' get yours, oh

In the middle of the night

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Step-by-step explanation:

5 0

2 years ago

nikitadnepr [17]

Answer:

$46

Step-by-step explanation:

40(0.15) = 6 (this is the amount of the tip)

40 + 6 = 46 (add the tip to the bill)

7 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