Solve for X. Please help ASAP! I have no idea how to do this and it’s due really soon!

If n erasers have a weight of 80 grams, what is the total weight of 50 erasers?

Black_prince [1.1K]

Answer:

$The\ weight\ of\ the\ 50\ erasers\ be\ \frac{4000}{n}\ grams.$

Step-by-step explanation:

As given

if n erasers have a weight of 80 grams .

I.e

n erasera = 80 grams.

1 erasers weight .

$1\ erasers\ weight = \frac{80}{n}$

Now find out the weight of the 50 erasers.

$Weight\ of\ 50\ erasers = \frac{50\times 80}{n}\ grams$

Simplify the above

$Weight\ of\ 50\ erasers = \frac{4000}{n}\ grams$

$Therefore\ the\ weight\ of\ the\ 50\ erasers\ be\ \frac{4000}{n}\ grams.$

6 0

3 years ago

Read 2 more answers

WL

otez555 [7]

The function appears to be L(legos) = T(tower)^3

L = T^3

This checks out for t =1,2,3,4

The 100th tower would have 100^3 legos.

100^3 = 1,000,000.

The 100th tower would have 1 million cubes

8 0

3 years ago

Of the 1000 students in a local college ,

dimaraw [331]

Can you please explain? Or maybe continue your question please?

5 0

2 years ago

How do make number lines

devlian [24]

Answer:

Select the plain line in the Insert Shapes group. Place your cursor where you want a tick mark on your line, hold the "Shift" key, and drag to create a vertical mark. Holding the Shift key ensures a straight line.

Step-by-step explanation:

5 0

3 years ago

<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:

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

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

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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

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

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