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Snezhnost [94]
3 years ago
13

Henry sells rings for $8 each. His expenses are $1.50 per ring, plus $91 for supplies. How many rings does he need to sell for h

is revenue to equal his expenses?
Mathematics
2 answers:
Firlakuza [10]3 years ago
8 0
Sorry, I’m not sure. Ask your teacher for help
kodGreya [7K]3 years ago
3 0

Answer:

14 rings.

Step-by-step explanation:

in order to calculate how many rings does he have to sell to break even we first have to calculate the profit that he has on the sell of every ring from which he can take money to pay per supplies, so that would be the selling price minues expenses:

8-1.5=6.5

Now we have to divide the 91 in supplies by the profit:

91/6.5=14

So he has to sell 14 rings in order to break even between expenses and supplies.

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HELP PLEASEEEEE I'll give you a crown
arlik [135]

Answer:

8.64%

Step-by-step explanation:

Write it as a decimal

7/81 = 0.0864

0.0864 is the decimal representation for 7/81

For Percentage Conversion :

step 1 To represent 0.0864 in percentage, write 0.0864 as a fraction

Fraction = 0.0864/1

step 2 multiply 100 to both numerator & denominator

(0.0864 x 100)/(1 x 100) = 8.64/100

8.64% is the percentage representation for 7/81

4 0
3 years ago
What are the square roots of 81? I​
ipn [44]
The square root of 81 is 9
7 0
2 years ago
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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
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frosja888 [35]

Answer:

In explanation

Step-by-step explanation:

272 for part A because say you took a test out of 850 and you got a total of 272 of them right then you would of gotten a 32%

200 for 2 because if 75% of them liked pop music and 150 teenagers what the number of teenagers surveyed then that means 200 of them were surveyed because 150 out of 200 is 75%

I don't know the last one because if you had anything there only could get a 100%(im sorry for this one it's a little confusing.

Hope this helps have a great afternoon:)

7 0
2 years ago
5th grade math, correct answer will be marked brainliest.
lina2011 [118]

Answer:

1.4 is the answer

Step-by-step explanation:

For every times 10 move the decimal point to the right.

3 0
3 years ago
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