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2 years ago

There are many answers not just one. I need help

Mathematics

2 answers:

Doss [256]2 years ago

8 0

Answer:

it's 3/4 feet

Step-by-step explanation:

serious [3.7K]2 years ago

3 0

Answer:

I think 1 and 5

Step-by-step explanation:

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A number is chosen at random from the positive, even integers from 2 to 50. Find the probability that the numbe

user100 [1]

Numbers divisible by 5 from 2 to 50 is 10 and numbers divisible by 8 from 2 to 50 is 6 ,therefore 6+ 10 =16 making 16/50 =8/25

8 0

2 years ago

An atom of silicon is approximately 0.00000000025 meters wide. How do you express that number in scientific notation? O 25 10-10

garik1379 [7]

Answer:

mohydty868788

Step-by-step explanation:

8 0

3 years ago

SOMEONE PUT AN ANSWER SO IT DOESNT GET DELETED!!!

Nadusha1986 [10]

Answer:

The height of the cylinder is 4x units.

Step-by-step explanation:

volume of a cylinder = nr^2h

n = 22/7

r = radius = x/2

Let illustrate with an example. If the diameter of a cylinder is 6 and the volume is 6^3 = 216. The radius = 6/2 = 3

n3^2 x h = 216

h = 216/9 = 24

so the height is 4 times the diameter

Area of the base = nr^2

1/2^2 x x = 1/4 pi x^2

3 0

2 years ago

Choose the best description for the real number square root of 35.

Liono4ka [1.6K]

$\sqrt{35}$

This cannot be simplified any further.

Have a good one!

3 0

2 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:

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

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

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

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

2 years ago