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Which of the following rational numbers are equivalent? $0Ptionsy$ A \frac{5}{6}, \frac{30}{36} B $s\sqrt{rac\left(} -2\right)\left(3\right)\sqrt{1rac} \sqrt{4\right)16\right)4} $ C $s\sqrt{11aC\left(} -4\right)1-7b,\sqrt{1rac\left(16\sqrt{35\right)9} } $ D \frac{1}{2},\frac{3}{8}

Can you answer this? $20$ $25$ $18$ $\left($ $\left(A\right)$ $A\right)2$ $21frac\left(5\right)\left(9\right)$ \) $\left(B\right)$ $B\right)$ $1\left(211$ $\left(C\right)$ $1\left(21$ $21+rac\left(7\right)+9\right)$ \) $\left(D\right)$ $1\left(2\right)$ 2\frac{8}{9} $ac\left(8\right)\left(9\right)$ \) $9:18PM\sqrt{} $

$\left(int|imits$ $-0a1\left(1-\times n_{2}{\right)_{3}}^{n}$ $x^{A}3dx=7\right)$ $\left($ $frac\left(1\right)\left(40\right)\right)$ $\left($ $\left(troc\left(1\right)\left(35\right)\right)$ $\left(troc\left(1\right)\left(30\right)\right)$ $\left(tr0c\left(1\right)\left(25\right)\right)$

In the following problem $a,b,$ b,and c represent REAL NUMBERS. The derivative of $g\left(x\right)=log _{a}\left(bx\right)+cx^{2}is$ $○$ $g^{1}\left(x\right)=\right)dfrac\left(b\right)log _{-}a\left(bx\right)\right)\left(N1n$ $a\right)+2cx1\right)$ $○$ $\left(g\left(x\right)=\right)dtracb$ $loga\left(bx\right)+2c\times \right)Nln$ a}\) $○$ $g^{'}\left(x\right)=\left(dfraC\left(a\right)log _{-}a\left(bx\right)\right)\left(ln$ $\right)+2c\times 1\right)\right)$ $○$ $g^{1}\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(ln$ $a1\right)$ $○$ $\left(\left(g^{1}\left(x\right)=b$ $log _{-}a\left(bx\right)+2cx1\right)$ $○$ $g\left(x\right)=\left(dfrac\left(b$ $log _{-}a\left(bx\right)\right)\left(1ln$ $a|+2c1\right)$

$1f\left(u=\left(rac\left(x^{n}3+y^{n}3\right)\left(x+y\right)\right)$ then $\left($ xfrac{partial $1\right)$ {partial $x\right)+y\left(raC\left(paF|a|$ u} {partial $y\right)=2\right)$ $0$ $0$ $20$ $301$

$11.$ Question $11$ Solve the $:$ $folloMlng'$ $0<θ<90^{°}$ $\left(1\right)$ $2sin^{2}θ=1\right)$ $\left(rac\left(3\right)\left(2\right)\right)$ $\left(11\right)$ $3tan^{2}θ+2=3$ $\left(111\right)cos^{2}θ$ $11rac\left(1\right)\left(4\right)\right)=$ $c\left(1\right)\left(4\right)\right)=11113c\left(1\right)\left(2\right)\right)$