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rationalize

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1
rationalize
rationalize, v. (ˈræʃənəlaɪz) [f. rational a.] 1. a. trans. To render conformable to reason; to explain on a rational basis.1803 Lett. Miss Riversdale II. 79 This interesting sentiment [sc. friendship]..secures the permanence of happiness, by rationalizing (if I may use such a word) its origin. 1817... Oxford English Dictionary
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rationalize
rationalizerationalise, / ˈræʃnəlaɪz; `ræʃənlˌaɪz/ v1 [I, Tn] (try to) justify (one's actions, emotions, etc) by giving a rational explanation for them （试图）使（自己的行动、 情感等）有合理依据 He's constantly rationalizing. 他总是不断反省是否言行合理. She rationalized her decision to abandon her baby by saying she could not affor... 牛津英汉双解词典
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4 Ways You Rationalize When You Act Against Your Conscience
Mar 8, 20223. Disregarding or Distorting the Consequences of Immoral Action. Another way of maintaining the morality of our self-concept is, selectively, to defocus from the effects of our action, whether ...
www.psychologytoday.com 0.0 0.90000004 0.0
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Cheaters Use Cognitive Tricks to Rationalize Infidelity
By coming up with these rationalizations, people are able to preserve the impression that their behaviors and attitudes are consistent. Similarly, cheaters might minimize the significance of their ...
www.scientificamerican.com 0.0 0.6 0.0
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Rationalize the denominator of the surd, giving your answer in the simplest form. Rationalize the denominator of the surd, giving your answer in the simplest form. $\frac {3}{\sqrt2+5} $ Please help me... It must b...
Yes, you've got a great start. Now, simply multiply the numerators and the denominators, $$\frac {3}{\sqrt2+5} * \frac{\sqrt2-5}{\sqrt2-5} = \dfrac{3(\sqrt 2 -5)}{(\sqrt 2 + 5)(\sqrt 2 - 5)}$$ - and in the denominator, use the fact that $$(a +b)(a-b) = a^2 - b^2$$
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Rationalize the denominator: $\frac {1}{\sqrt[3]{x}+2}$ How am I supposed to rationalize the denominator for $\frac {1}{\sqrt[3]{x}+2}$? I don't even know where to really begin. I tried multiplying both the numerator...
You will have to multiply both the numerator and the denominator by sums of cubes. Sums of Cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$ So to get $\sqrt[3]{x}+2$ into a perfect number with no radicals, you have to multiply it by $\sqrt[3]{x^2}-2\sqrt[3]{x}+\sqrt[3]{64}$. Doing so to the numerator and denomina...
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Rationalize Below Equation How can I rationalize the following equation: $$\frac{22}{4\sqrt[3]{9}+2\sqrt[3]{6}+\sqrt[3]{4}}$$
Let $a=2\sqrt[3]{3}, b=\sqrt[3]{2}$. Then this is equal to $\dfrac{22}{a^2+ab+b^2} = \dfrac{22(a-b)}{a^3-b^3}$. Since $a^3-b^3 = 8(3)-2 =22$, we have that our desired answer is $\dfrac{22(a-b)}{a^3-b^3} = a-b = 2\sqrt[3]{3}-\sqrt[3]{2}$.
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Rationalize and evaluate when $h=0: \frac{\sqrt{16+h}-4}{h}$ Rationalize and evaluate when $h=0: \frac{\sqrt{16+h}-4}{h}$ I know how to rationalize when there is a radical in the denominator (multiply by conjugate) b...
$$\begin{align} \frac{\sqrt{16+h}-4}{h} &= \frac{\sqrt{16+h}-4}{h}\left(\dfrac{\sqrt {16+h}+4}{\sqrt {16+h}+4}\right) =\dotsb\\\\\\\\\dotsb&= \frac{(16+h)-16}{h(\sqrt {16+h}+4)} = \frac{1}{\sqrt {16+h}+4} \end{align}$$
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he li hua : to rationalize,... : hé lǐ huà | Definition | Mandarin ...
he li hua definition at Chinese.Yabla.com, a free online dictionary with English, Mandarin Chinese, Pinyin, Strokes & Audio. Look it up now!
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Rationalize a surd $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ How can I rationalize the following surd $$\frac{1}{1+\sqrt{2}-\sqrt{3}}$$ What would be the conjugate of the denominator
Rationalise twice, because after rationalising once , there would still remain a surd in the denominator. First multiply and divide by $1+\sqrt{2}+ \sqrt{3}$ $$\frac{1}{1+\sqrt{2}-\sqrt{3}}\times\frac{1+\sqrt{2}+\sqrt{3}}{1+\sqrt{2}+\sqrt{3}}$$ $$= \frac{1+\sqrt{2}+\sqrt{3}}{1+2\sqrt{2}+2-3}=\frac{1...
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How to rationalize this root form? Suppose that we have a equation like this: $$\sqrt{a+b+2\sqrt{ab}}$$ or $$\sqrt{a+b-2\sqrt{ab}}$$ In order to rationalize it, we can apply the formula: $$\sqrt{a} + \sqrt{b} = \sq...
Notice that there are $3$ terms with $2$ so probably it will be of form $(a+b+c)^2$ $$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\\\a^2+b^2+c^2=10\\\ab=\sqrt{15}\\\ac=\sqrt{10}\\\bc=\sqrt{6}\\\b^2+c^2=10-a^2\\\a^2b^2+a^2c^2=25\\\a^2(b^2+c^2)=25\\\a^2(10-a^2)=25\\\10a^2-a^4-25=0\\\a^4-10a^2+25=0\\\\(a^2-5)^2=0...
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Why we rationalize, conjugate. My question is why we rationalise, conjugate any denominator containing irrational or imaginary quantity. What is the need to rationalize them?
As a tool to solve problems, sometimes it is desirable to do so. For instance, $$\lim_{x\rightarrow 0} \frac x{\sqrt{x+1}-1} =\lim_{x\rightarrow 0} \frac {x(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)(\sqrt{x+1}+1)} =\lim_{x\rightarrow 0} \frac {x(\sqrt{x+1}+1)}{x} =\lim_{x\rightarrow 0} \sqrt{x+1}+1 =2$$
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How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt...
We have $(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=x^3+y^3+z^3-3xyz$ With $x=9\sqrt[3]{9},y=-3\sqrt[3]{3},z=-27$ all terms on the right side are rational, try it. So multiply the given numerator and denominator by $x^2+y^2+z^2-xy-xz-yz$ with $x,y,z$ as rendered above.
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How do I rationalize the equation $\frac{3-x}{1- \sqrt{x-2}}$? How do I rationalize this equation $$\frac{3-x}{1- \sqrt{x-2}} ? $$
$$\frac{3-x}{1-\sqrt{x-2}}=\frac{3-x}{1-\sqrt{x-2}}\frac{1+\sqrt{x-2}}{1+\sqrt{x-2}}=\frac{(3-x)(1+\sqrt{x-2})}{1-(x-2)}=$$ $$=\frac{(3-x)(1+\sqrt{x-2})}{3-x}=1+\sqrt{x-2}$$
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Rationalize the numerator of each of the following expressions √a - 2 / a -4 the answer is supposed to be 1 / √a + 2 Please show me the steps of how you got this answer please
So $a - 4 = (\sqrt a - 2)(\sqrt a + 2)$, giving you a fraction of $$ \frac{\sqrt a - 2}{ (\sqrt a - 2)(\sqrt a + 2)} = \frac{1}{\sqrt a + 2}$$ as desired. This turns out to be a problem where you recognize that the denominator can be factored by something in the numerator - basically it's a pattern ...
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