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3a^{2}
Odvajajte w.r.t. a
6a
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Algebra
\sqrt{3} \times \sqrt{3a^4}
Podobne težave pri spletnem iskanju
Simplify? \displaystyle\sqrt{{8}}\times\sqrt{{{48}^{{3}}}}
https://socratic.org/questions/59e559a97c01496bf2104ce3
\displaystyle\sqrt{{8}}\times\sqrt{{{48}^{{3}}}}={384}\sqrt{{6}} Explanation: \displaystyle\sqrt{{8}}\times\sqrt{{{48}^{{3}}}} Because both terms are under a square root sign, we can ...
How do you simplify \displaystyle{5}\sqrt{{{9}{t}^{{2}}}}\times{5}\sqrt{{{2}{t}}} ?
https://socratic.org/questions/how-do-you-simplify-5sqrt-9t-2-times5-sqrt-2t
See a solution process below: Explanation: First, simplify the radical on the left: \displaystyle{\left({5}\times{3}{t}\right)}\times{5}\sqrt{{{2}{t}}}\Rightarrow \displaystyle{15}{t}\times{5}\sqrt{{{2}{t}}}\Rightarrow ...
How do you simplify \displaystyle{3}\sqrt{{{5}{c}}}\times\sqrt{{15}}^{{3}} ?
https://socratic.org/questions/how-do-you-simplify-3sqrt-5c-times-sqrt15-3
\displaystyle{225}\sqrt{{{3}{c}}} Explanation: \displaystyle{3}\sqrt{{{5}{c}}}\sqrt{{{15}}}^{{3}} First, we can simplify \displaystyle\sqrt{{{15}}}^{{3}} . \displaystyle\sqrt{{{15}}}^{{3}}=\sqrt{{15}}\cdot\sqrt{{15}}\cdot\sqrt{{15}}={15}\cdot\sqrt{{15}} ...
Simplifying indices with surds
https://math.stackexchange.com/questions/1986172/simplifying-indices-with-surds
One way is to note that \left( \sqrt t \right)^3=t^{\frac 32} and similarly for the other one. Then when you multiply terms you add exponents
range of m such that the equation |x^2-3x+2|=mx has 4 real answers.
https://math.stackexchange.com/questions/1259271/range-of-m-such-that-the-equation-x2-3x2-mx-has-4-real-answers
There is some positive value m such that y=mx is tangent to y=-(x^2-3x+2). This value must make 0 the discriminant of the equation x^2-3x+2=-mx That is, m^2-6m+1=0 The least root of ...
Prove that there exists irrational numbers p and q such that p^{q} is rational
https://math.stackexchange.com/q/2883337
The irrationality of \sqrt 2^{\sqrt 2} (in fact, its transcendence) follows immediately from the Gelfond Schneider Theorem . This was the issue that motivated Hilbert's 7^{th} Problem. The ...
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\sqrt{40}
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\sqrt{3} \times \sqrt{3a^4}
\sqrt{\sqrt{256a^8}}
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