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ycow [4]
3 years ago
5

There are 2 miners in a mine. Miner 1 mines 19,687 m of ore in one hour. Miner 2 mines 19,687 m of ore in 51 minutes. Find out h

ow much ore Miner 2 mines in 1 hour.
Mathematics
1 answer:
patriot [66]3 years ago
8 0
Miner #2 mines 19,687 m (meters ?) of ore in 51 minutes.

That's  (19,687/ 51) m (meters ?) per minute.

There are 60 minutes in one hour.  So if he maintains that same
average rate for a whole hour, he will mine

           (19,687 / 51) x (60)   =   23,161 m (meters) of ore.
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<img src="https://tex.z-dn.net/?f=%5Csf%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Ccfrac%7B%5Csqrt%7Bx-1%7D-2x%20%7D%7Bx-7%7D" id=
BARSIC [14]
<h3>Answer:  -2</h3>

======================================================

Work Shown:

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}\left(\sqrt{x-1}-2x\right) }{ \frac{1}{x}\left(x-7\right) }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \frac{1}{x}*\sqrt{x-1}-\frac{1}{x}*2x }{ \frac{1}{x}*x-\frac{1}{x}*7 }\\\\\\

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}}*\sqrt{x-1}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x^2}*(x-1)}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{\frac{1}{x}-\frac{1}{x^2}}-2 }{ 1-\frac{7}{x} }\\\\\\\displaystyle L = \frac{ \sqrt{0-0}-2 }{ 1-0 }\\\\\\\displaystyle L = \frac{-2}{1}\\\\\\\displaystyle L = -2\\\\\\

-------------------

Explanation:

In the second step, I multiplied top and bottom by 1/x. This divides every term by x. Doing this leaves us with various inner fractions that have the variable in the denominator. Those inner fractions approach 0 as x approaches infinity.

I'm using the rule that

\displaystyle \lim_{x\to\infty} \frac{1}{x^k} = 0\\\\\\

where k is some positive real number constant.

Using that rule will simplify the expression greatly to leave us with -2/1 or simply -2 as the answer.

In a sense, the leading terms of the numerator and denominator are -2x and x respectively. They are the largest terms for each, so to speak. As x gets larger, the influence that -2x and x have will greatly diminish the influence of the other terms.

This effectively means,

\displaystyle L = \lim_{x\to\infty} \frac{ \sqrt{x-1}-2x }{ x-7 } = \lim_{x\to\infty} \frac{ -2x }{ x} = -2\\\\\\

I recommend making a table of values to see what's going on. Or you can graph the given function to see that it slowly approaches y = -2. Keep in mind that it won't actually reach y = -2 itself.

5 0
3 years ago
Can someone help, please?!
sergiy2304 [10]

Answer:

one solution

Step-by-step explanation:

4 0
3 years ago
Given cosθ=√3/3 and sinθ&lt;0. What is the value of sinθ?
KATRIN_1 [288]

\sin^{2} \theta+\cos^{2} \theta=1 \\ \\ \sin^{2} \theta+\left(\frac{\sqrt{3}}{3} \right)^{2}=1 \\ \\ \sin^{2} \theta+\frac{1}{3}=1 \\ \\ \sin^{2} \theta=\frac{2}{3} \\ \\ \theta=-\frac{\sqrt{2}}{\sqrt{3}}=\boxed{-\frac{\sqrt{6}}{3}}

5 0
2 years ago
How do you convert a whole number to a fraction?
Maslowich

Answer:

Step-by-step explanation:

Example 1: Changing the whole number 5 into a fraction.

Take the whole number (5), add a line below it (/), then add a 1 to the denominator.

5 = 5/1

Example 2: Changing the whole number 5 into a fraction.

Take the whole number (5), multiply it by 2 add a line below it (/), then add a 2 to the denominator.

5 = (5*2)/2 = 10/2

***This can be reduced to 5/1***

Example 3: Changing the whole number 5 into a fraction.

Take the whole number (5), multiply it by 3 add a line below it (/), then add a 3 to the denominator.

5 = (5*3)/3 = 15/3

***This can also be reduced to 5/1***

If you follow the pattern, you will realize all whole numbers are fractions already.

They are fractions with a denominator of 1. This fraction can be manipulated with all of the same standard rules you would traditionally use with fractions, even when the denominator isn’t shown.

A fraction is simply a way to describe portions of a whole. The denominator simply tells you how many pieces to break the whole into. When the denominator is 1, you are breaking the whole into one piece (or not breaking it apart at all.

I hope this helps.

8 0
3 years ago
What is the greatest common factor of the terms in the polynomial 4x^4-32x^3-60x^2?
jonny [76]

Answer:

\mathrm{Factor}\:4x^4-32x^3-60x^2:\quad 4x^2\left(x^2-8x-15\right)\\

\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\x^3=xx^2\\x^4=x^2x^2\\=4x^2x^2-32xx^2-60x^2\\\mathrm{Rewrite\:}60\mathrm{\:as\:}4\cdot \:15\\\mathrm{Rewrite\:}32\mathrm{\:as\:}4\cdot \:8\\=4x^2x^2-4\cdot \:8xx^2-4\cdot \:15x^2\\\mathrm{Factor\:out\:common\:term\:}4x^2\\=4x^2\left(x^2-8x-15\right)

<em>Hope this helps and have a great day!!!</em>

<em>Sofia</em>

4 0
3 years ago
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