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Galina-37 [17]
3 years ago
14

List the term of the expression. mn-2n+8

Mathematics
1 answer:
ELEN [110]3 years ago
6 0
I'm assuming you mean to list the terms of the expression.  Those are mn, -2n, and +8.  There are 3 terms there.
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When you multiply 2 decimals, how do you know where to place the comma in the product?
sladkih [1.3K]

Answer:

you mean the decimal point?

Step-by-step explanation:

You will know when you cross the decimal point when multiplying.

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From an electronic shop, 34 washing machines were sold on a particular day. The price of one washing machine was 8,400. Find the
Mamont248 [21]

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$201,600

Step-by-step explanation:

Multiply 34 and 84, which is 2016. Then add 2 zeros at the end because there are 2 zeros in 8400. So, the answer is 201,600 dollars.

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Step-by-step explanation:

8 0
3 years ago
Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→1 3x x − 1
Reika [66]

Answer:

<h2>3/2</h2>

Step-by-step explanation:

Given the limit of a function expressed as \lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx}), we are to evaluate it. To evaluate it, we will simply substitute x = 1 into the function since the variable x tends to 1.

\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= (\dfrac{3(1)}{1-1} - \dfrac{3}{ln1})\\\\= \dfrac{3}{0} - \dfrac{3}{0}\\\\= \infty - \infty (ind)

Since we got an indeterminate function, we will find the LCM of the function and solve again.

= \lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= \lim_{x \to 1} \dfrac{3xlnx-3(x-1)}{(x-1)lnx}\\\\\\= \dfrac{3(1)ln(1)-3(1-1)}{(1-1)ln1}\\\\= \frac{3(0)-3(0)}{0(0)} \\\\= \frac{0}{0} (ind)

Applying L'hospital rule;

\frac{x}{y} = \lim_{x \to 1} \dfrac{d/dx(3xlnx-3(x-1))}{d/dx((x-1)lnx)}\\\\=  \lim_{x \to 1} \dfrac{3x(\frac{1}{x})+ 3lnx-3)}{(x-1)\frac{1}{x} +lnx}\\\\= \lim_{x \to 1} \dfrac{3 + 3lnx-3}{(x-1)\frac{1}{x} +lnx}\\\\= \frac{3ln1}{(1-1)\frac{1}{1} +ln1}\\\\= \frac{0}{0} (ind)

Applying L'hospital rule again;

= \lim_{x \to 1} \dfrac{\frac{d}{dx} (3lnx)}{\frac{d}{dx} ((x-1)\frac{1}{x} +lnx)}\\\\=  \lim_{x \to 1} \dfrac{\frac{3}{x} }{(x-1)\frac{-1}{x^2} + \frac{1}{x} +\frac{1}{x} }\\\\= \dfrac{\frac{3}{1} }{(1-1)\frac{-1}{1^2} + \frac{1}{1} +\frac{1}{1} }\\\\= \frac{3}{0(-1)+2}\\ \\= \frac{3}{2-0}\\ \\= 3/2

<em>Hence the limit of the function is 3/2.</em>

7 0
4 years ago
What is the interquartile range of this set of data? 15, 19, 20, 25, 31, 38, 41.
Korvikt [17]
The answer is D, 26.
7 0
3 years ago
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