Answer:
i think it is 8/125
Step-by-step explanation:
slect B
Answer:
x = -3
y = 0
Step-by-step explanation:
<u>Given</u><u> </u><u>equations</u><u> </u><u>:</u><u>-</u><u> </u>
<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u>
<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u>
<u>From</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u><u> </u><u> </u>
<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u>
<u>-x</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>-</u><u> </u><u>2</u><u>y</u><u> </u>
<u>x</u><u> </u><u>=</u><u> </u><u>2</u><u>y</u><u> </u><u>-</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u>
<u>From</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u><u> </u>
<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u>
<u>2</u><u>x</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>+</u><u> </u><u>3</u><u>y</u><u> </u>
<u>
</u>
<u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>
<u>Equating</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>
<u>x</u><u> </u><u>=</u><u> </u><u>x</u><u> </u>
<u>
</u>
4y - 6 = -6 + 3y
4y - 3y = -6 + 6
y = 0
Putting value of y in ( iii )
x = 2y - 3
x = 2 ( 0 ) - 3
x = -3
Answer:
trapezium................
<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 }\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B%20%5Csqrt%7Bx-1%7D-2x%20%7D%7B%20x-7%20%7D%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B%20%5Cfrac%7B1%7D%7Bx%7D%5Cleft%28%5Csqrt%7Bx-1%7D-2x%5Cright%29%20%7D%7B%20%5Cfrac%7B1%7D%7Bx%7D%5Cleft%28x-7%5Cright%29%20%7D%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B%20%5Cfrac%7B1%7D%7Bx%7D%2A%5Csqrt%7Bx-1%7D-%5Cfrac%7B1%7D%7Bx%7D%2A2x%20%7D%7B%20%5Cfrac%7B1%7D%7Bx%7D%2Ax-%5Cfrac%7B1%7D%7Bx%7D%2A7%20%7D%5C%5C%5C%5C%5C%5C)
![\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\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B%20%5Csqrt%7B%5Cfrac%7B1%7D%7Bx%5E2%7D%7D%2A%5Csqrt%7Bx-1%7D-2%20%7D%7B%201-%5Cfrac%7B7%7D%7Bx%7D%20%7D%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B%20%5Csqrt%7B%5Cfrac%7B1%7D%7Bx%5E2%7D%2A%28x-1%29%7D-2%20%7D%7B%201-%5Cfrac%7B7%7D%7Bx%7D%20%7D%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B%20%5Csqrt%7B%5Cfrac%7B1%7D%7Bx%7D-%5Cfrac%7B1%7D%7Bx%5E2%7D%7D-2%20%7D%7B%201-%5Cfrac%7B7%7D%7Bx%7D%20%7D%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20L%20%3D%20%5Cfrac%7B%20%5Csqrt%7B0-0%7D-2%20%7D%7B%201-0%20%7D%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20L%20%3D%20%5Cfrac%7B-2%7D%7B1%7D%5C%5C%5C%5C%5C%5C%5Cdisplaystyle%20L%20%3D%20-2%5C%5C%5C%5C%5C%5C)
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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\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B1%7D%7Bx%5Ek%7D%20%3D%200%5C%5C%5C%5C%5C%5C)
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\\\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20L%20%3D%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B%20%5Csqrt%7Bx-1%7D-2x%20%7D%7B%20x-7%20%7D%20%3D%20%5Clim_%7Bx%5Cto%5Cinfty%7D%20%5Cfrac%7B%20-2x%20%7D%7B%20x%7D%20%3D%20-2%5C%5C%5C%5C%5C%5C)
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.
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
The value 5 represent in this situation is letter B which is "
<span>The value of the quarters in the bowl on Week 1 was $5."</span>