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lukranit [14]
4 years ago
14

The sum of the digits of a two-digit number is 5. If nine is subtracted from the number, the digits will be reversed. Find the n

umber. the answer to that is 32. If you replace the tens digit with x, then what will the equation be?
Mathematics
1 answer:
Sunny_sXe [5.5K]4 years ago
5 0

Answer: 32

Step-by-step explanation:

Let x + y = 5

x = 5 - y....................(1)

10x + y -9 = 10y + x..... . (2)

Substituting (1) in (2)

10(5-y) + y - 9 = 10y + (5-y)

50-10y + y - 9 = 10y + 5 - y

41 - 9y = 9y + 5

-9y - 9y = 5-41

-18y = - 36

y = 36/2 = 2

So, x = 5-2 = 3

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Answer:

x= 175°

Step-by-step explanation:

angles in a triangle add up to 180 degrees

85 + 90 = 175

180 - 175 = 5 (the missing angle not x)

angles on a straight line add up to 180 degrees

180 -  5 =175

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A fruit company delivers its fruit in two types of boxes: large and small. A delivery of 3 large boxes and 5 small boxes has a t
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Answer:

You can not tell unless you have information about the relationship between the small and large boxes (for example, a large box is 2 times the weight of a small box) You can only tell that a small box and a large box's weight together is 34.5 kilograms.

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3L + 5S = 135

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aleksandrvk [35]
Yo what is the answer??
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I need help with this problem from the calculus portion on my ACT prep guide
LenaWriter [7]

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):

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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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