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3 years ago

Someone please help!!

Mathematics

2 answers:

marta [7]3 years ago

8 0

Answer:

Van: 18, Bus: 59

Step-by-step explanation:

b - number of students that bus can carry

v - number of students that van can carry

6•b + 1•v = 372 - students from A

6b + v = 372 ⇒ v = 372 - 6b

12•b + 4•v = 780 - students from B

12b + 4(372 - 6b) = 780

12b + 1488 - 24b = 780

- 12x = - 708

Katena32 [7]3 years ago

5 0

After solving for both variables, you find that each bus can hold 59 students and each van can hold 18 students.

Step-by-step explanation:

You can find the amount of students each vehicle can carry by representing the two scenarios in equations.

You are trying to find how many students will fit in each bus or van, so the two variables used will be "b" to represent how many students can fit in a bus and "v" to represent how many students can fit in a van.

High school A used 1 van and 6 buses, so there will be 1"v" and 6"b" for 372 students.

High school B used 4 vans and 12 buses, so there will be 4"v" and 12"b" for 780 students.

Now, represent these in equations:

$v+6b = 372\\4v + 12b = 780$

We can use substitution to solve this system:

$v + 6b = 372$ can be rewritten as $v = 372 - 6b$ after subtracting 6b from both sides. You can then substitute this new value of "v" into the other equation to solve for "b":

$4(372-6b) + 12b = 780\\1488 - 24b + 12b = 780\\1488 - 12b = 780\\-12b = -708\\b = 59$

After solving for b, you can then substitute the new value of b into the other equation to find the value of v:

$v + 6(59) = 372\\v + 354 = 372\\v = 18$

After solving for both variables, you find that each bus can hold 59 students and each van can hold 18 students.

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Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

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2 years ago