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3 years ago

16. You want to decorate around the top of a jar with ribbon. The length around the jar is

Mathematics

1 answer:

VladimirAG [237]3 years ago

3 0

Answer: A) We need 1 1/2 ft of ribbon

B)$1.60 per foot

C)$2.40 is the total

Step-by-step explanation:

So if the length around the jar is 18 inches and we know that there is 12 inches in a foot, and 3 feet in a yard and $ 4.80 per every 3 feet. These become basis numbers. 18in- 12inch= 6 inch. So now we have 1 1/2 feet of ribbon to cover the circumference of the jar. Now we are left to answer the second question how much does it cost per foot? We can divide 4.80 by 3 to see the total cost of a foot of ribbon. That turns to 1.60 per foot. We divide that by two to cover the 6 inches. And the answer is $2.40 for 1 1/2 feet of ribbon

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2 years ago

Five guys walk into a bar, how many ways can they sit so as to be arranged from oldest to youngest?

Ipatiy [6.2K]

There are 24 ways in which 5 guys can sit if arranged from oldest to youngest.

We have,

Five guys.

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We know that,

Total number of ways to arranged around a table (n) = (n-1)!

So,

For n = 5,

I.e.

(n - 1)! = (5 - 1)! = 4!

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Total number of ways to arranged Five guys rom oldest to youngest (n) = (n-1)!

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Total number of ways to arranged Five guys rom oldest to youngest (n) = 24.

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1 year ago

The half-life of a radioactive element is 130 days, but your sample will not be useful to you after 80% of the radioactive nu

gtnhenbr [62]

Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt} =-\lambda N$

$\Rightarrow \frac{dN}{N} =-\lambda dt$

Integrating both sides

$\int \frac{dN}{N} =\int\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$\Rightarrow ln|N_0|=-\lambda .0+c$

$\Rightarrow c= ln|N_0|$

$\therefore ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$ .......(1)

$\frac{N}{N_0}=e^{-\lambda t}$ .........(2)

Logarithm:

$ln|\frac mn|= ln|m|-ln|n|$
$ln|ab|=ln|a|+ln|b|$

$ln|e^a|=a$

$ln|a|=b \Rightarrow a=e^b$
$ln|1|=0$

130 days is the half-life of the given radioactive element.

For half life,

$N=\frac12 N_0$ , $t=t_\frac12=130$ days.

we plug all values in equation (1)

$ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130$

$\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130$

$\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130$

$\rightarrow -ln|2|=-\lambda \times 130$

$\rightarrow \lambda= \frac{-ln|2|}{-130}$

$\rightarrow \lambda= \frac{ln|2|}{130}$

We need to find the time when the sample remains 80% of its original.

$N=\frac{80}{100}N_0$

$\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t$

$\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t$

$\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}$

$\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}$

$\Rightarrow t\approx 42$

We can use the sample about 42 days.

7 0

3 years ago