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Andre45 [30]
1 year ago
10

A hydrogen atom has one positively charged proton and one negatively charged electron.

Mathematics
2 answers:
vladimir1956 [14]1 year ago
7 0

Answer:

i don't know the answer yet , I'm quite not sure

charle [14.2K]1 year ago
3 0
Same it is quite confusing
You might be interested in
<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7
salantis [7]

Answer:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}

General Formulas and Concepts:

<u>Calculus</u>

Limits

Limit Rule [Variable Direct Substitution]:                                                                     \displaystyle \lim_{x \to c} x = c

L'Hopital's Rule

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

We are given the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

\displaystyle  \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle  \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}

Plugging in <em>x</em> = 0 again, we would get:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}

Substitute in <em>x</em> = 0 once more:

\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0
3 years ago
3
kari74 [83]

The equations to represent the relationship are ƒx3/5=4 4/5 and 4 4/5 divided 3/5 =f

<h3>Ratio and proportion</h3>

Fractions are written as a ratio of two integers. For instance a/b is a fraction.

If Anastacio distributed 4 4/5 kg of seed among several bird feeders and put 3/5 kg of seed in each feeder, the expression that represents the number of bird feeders that Anastacio filled is given as;

Number of bird feeder = 4 4/5 ÷ 3/5

f = 4 4/5 ÷ 3/5

Multiply both sides by

f * 3/5 = 4 4/5   ÷ 3/5 * 3/5

f * 3/5 = 4 4/5

Hence the equations to represent the relationship are ƒx3/5=4 4/5 and 4 4/5 divided 3/5 =f

Learn more on division of fraction here: brainly.com/question/407943

#SPJ1

3 0
1 year ago
Plz someone help me with this problem
fiasKO [112]
Not factorable
hajajaksmsmnsnsnsnsjd
6 0
3 years ago
2+99393×17888÷2=? <br> WHATS THE ANSWER DAWG
Anna35 [415]

Answer:

888970994

Step-by-step explanation:

99393 * 17888 = 1777941984

1777941984/2 = 888970992

888970992+2 = 888970994

3 0
3 years ago
Read 2 more answers
two nearby towns have a population of 442,680 and 564,943. what is the total population of both towns
stellarik [79]
1007623 is the answer because 442680 plus 564943 is 1007623
4 0
3 years ago
Read 2 more answers
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