4.

Mathematics

1 answer:

NemiM [27]3 years ago

6 0

Answer:

i belive the answer is B

Step-by-step explanation:

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(60y – 18) = 2(8-16)

Alekssandra [29.7K]

Answer:

3/10

Step-by-step explanation:

(60y-18)=2(8-16)

60y-18=16-16

60y-18=0

60y=18 /:60

y= 18/60=3/10

4 0

2 years ago

How many arrangements can be made using 2 letters of the word HYPERBOLAS if no letter is to be used more than once?

alisha [4.7K]

Answer:

90 arrangements

Step-by-step explanation:

Since there are no repititions of letters, there are unique 10 letters in total.

THe number of arrangements would be 2 permutation 10. We need the formula for permutation. That is:

$nPr=\frac{n!}{(n-r)!}\\$

Now, n = 10 [total] and r is 2, so we have:

$nPr=\frac{n!}{(n-r)!}\\\\_{10}P_{2}=\frac{10!}{(10-2)!}\\=\frac{10!}{8!}\\=\frac{10*9*8!}{8!}\\=10*9\\=90$

So, there can be 90 arrangements

3 0

3 years ago

Gina puts $ 4500 into an account earning 7.5% interest compounded continuously. How long will it take for the amount in the acco

Elza [17]

$~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}$

$5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}$