Rezolvați 1/a^2+1/b^2=1/h^2 | Microsoft Math Solver

Rezolvați pentru a (complex solution)

a=-i\left(h^{2}-b^{2}\right)^{-\frac{1}{2}}bh

a=i\left(h^{2}-b^{2}\right)^{-\frac{1}{2}}bh\text{, }b\neq 0\text{ and }h\neq 0\text{ and }b\neq -h\text{ and }b\neq h

Rezolvați pentru b (complex solution)

b=-i\left(h^{2}-a^{2}\right)^{-\frac{1}{2}}ah

b=i\left(h^{2}-a^{2}\right)^{-\frac{1}{2}}ah\text{, }a\neq 0\text{ and }h\neq 0\text{ and }a\neq -h\text{ and }a\neq h

Rezolvați pentru a

a=\sqrt{-\frac{1}{h^{2}-b^{2}}}bh

a=-\sqrt{-\frac{1}{h^{2}-b^{2}}}bh\text{, }h\neq 0\text{ and }b\neq 0\text{ and }|b|>|h|

Rezolvați pentru b

b=\sqrt{-\frac{1}{h^{2}-a^{2}}}ah

b=-\sqrt{-\frac{1}{h^{2}-a^{2}}}ah\text{, }h\neq 0\text{ and }a\neq 0\text{ and }|a|>|h|

Test

Algebra

5 probleme similare cu aceasta:

\frac { 1 } { a ^ { 2 } } + \frac { 1 } { b ^ { 2 } } = \frac { 1 } { h ^ { 2 } }

Probleme similare din căutarea web

If the circles \displaystyle{x}^{{2}}+{y}^{{2}}+{2}{a}{x}+{c}^{{2}}={0} and \displaystyle{x}^{{2}}+{y}^{{2}}+{2}{b}{y}+{c}^{{2}}={0} touch externally prove that \displaystyle\frac{{1}}{{{a}^{{2}}}}+\frac{{1}}{{{b}^{{2}}}}=\frac{{1}}{{{c}^{{2}}}} ...

https://socratic.org/questions/if-the-circles-x-2-y-2-2ax-c-2-0-and-x-2-y-2-2by-c-2-0-touch-externally-prove-th

P dilip_k Nov 25, 2017 If the circles \displaystyle{x}^{{2}}+{y}^{{2}}+{2}{a}{x}+{c}^{{2}}={0} and \displaystyle{x}^{{2}}+{y}^{{2}}+{2}{b}{y}+{c}^{{2}}={0} touch externally prove ...

If two circles \displaystyle{x}^{{2}}+{y}^{{2}}+{2}{a}{x}+{c}^{{2}}={0} and \displaystyle{x}^{{2}}+{y}^{{2}}+{2}{b}{y}+{c}^{{2}}={0} touch each other externally then prove that \displaystyle\frac{{1}}{{{a}^{{2}}}}+\frac{{1}}{{{b}^{{2}}}}=\frac{{1}}{{{c}^{{4}}}} ...

https://socratic.org/questions/if-two-circles-x-2-y-2-2ax-c-2-0-and-x-2-y-2-2by-c-2-0-touch-each-other-external

Please refer to a Proof in Explanation. Explanation: Let the given circles be, \displaystyle{S}_{{1}}:{x}^{{2}}+{y}^{{2}}+{2}{a}{x}+{c}^{{2}}={0} . \displaystyle\therefore{\left({x}+{a}\right)}^{{2}}+{y}^{{2}}={\left(\sqrt{{{a}^{{2}}-{c}^{{2}}}}\right)}^{{2}},{\left({a}^{{2}}-{c}^{{2}}\right)}{>}{0} ...

What is the correct order for \displaystyle-{1}{\frac{{{2}}}{{{5}}}},{\left(-{1}\right)}^{{{2}}},-{1.4}{\quad\text{and}\quad}{\left({\frac{{{1}}}{{{2}}}}^{{2}}\right)} ?

https://socratic.org/questions/what-is-the-correct-order-for-1-frac-2-5-1-2-1-4-and-frac-1-2-2

\displaystyle-{1}\frac{{2}}{{5}},\ \text{ }\ -{1.4},\ \text{ }\ \frac{{1}^{{2}}}{{2}},\ \text{ }\ {1} Explanation: Convert them into the same form. Decimals are easiest to compare. \displaystyle-{1}\frac{{2}}{{5}},\ \text{ }\ {\left(-{1}\right)}^{{2}},\ \text{ }\ -{1.4},\ \text{ }\ \frac{{{1}^{{2}}}}{{2}} ...

(1/a^2)+-(1/b^2)/(1/ab)

https://www.tiger-algebra.com/drill/(1/a~2)_-(1/b~2)/(1/ab)/

(1/a2)+-(1/b2)/(1/ab) Final result : (b - a) • (a2 + ab + b2) ———————————————————————— a2b3 Step by step solution : Step 1 : 1 Simplify — a Equation at the end of step 1 : 1 1 1 ————-———— ÷ ...

The height of the tide measured at a seaside community varies according to the number of hours t after midnight. If the height h, in feet, is currently given by the equation \displaystyle{h}=-\frac{{1}}{{2}}{t}^{{2}}+{6}{t}-{9} ...

https://socratic.org/questions/the-height-of-the-tide-measured-at-a-seaside-community-varies-according-to-the-n

At \displaystyle{8.27} a.m. or \displaystyle{08.27} Explanation: Putting the value of h = 6 in equation \displaystyle{h}=-\frac{{1}}{{2}}{t}^{{2}}+{6}{t}-{9} or, \displaystyle{6}=\frac{{-{t}^{{2}}+{12}{t}-{18}}}{{2}} ...

find a polynomial whose roots are inverse of squares of roots of x^3+px+q

https://math.stackexchange.com/questions/872264/find-a-polynomial-whose-roots-are-inverse-of-squares-of-roots-of-x3pxq

x^3+px-q has roots -a,-b,-c. (x^3+px+q)(x^3+px-q) has roots a,b,c,-a,-b,-c. x^3+2px^2+p^2x-q^2 has roots a^2,b^2,c^2. 1+2px+p^2x^2-q^2x^3 has roots 1/a^2,1/b^2,1/c^2.

Partajați

Copiat în clipboard

Exemple

Ecuație de gradul 2

{ x } ^ { 2 } - 4 x - 5 = 0

Trigonometrie

4 \sin \theta \cos \theta = 2 \sin \theta

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

Sistem de ecuații

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

Derivare

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

Integrare

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

Limite

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

Subiecte

Subiecte

Probleme similare din căutarea web

Partajați

Exemple