Question

need someone's help urgently

Answer 1

$\textbf{ Hello Mate! }$

$1) \: {81}^{0.16} \times {81}^{0.09} \\ \\ = {81}^{0.16 + 0.09} = {81}^{0.25 } \\ \\ = {81}^{ \frac{25}{100} } = { 3 }^{4\times \frac{1}{4} } = 3$

$2) \: (2 + 3x)(2 - 3x) \\ \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ {2}^{2} - {(3x)}^{2} = 4 - 9 {x}^{2}$

3) <ACD = <A + <B [ Exterior angle property ]

<ACD = 40° + 70°

<ACD = 110°

Now, <ACD = <BCE = 110°

$\boxed{ \textsf{Answer= 1)\: 3 }}$

$\boxed{ \textsf{ 2)\: 4 - 9x sq }}$

$\boxed{ \textsf{3)\: Angle\:BCE = 110° }}$

Have great future ahead!

Answer 2

$\bf \large \it{Hey \: User!!!}$

$\rm 1.\: given : - {(81)}^{0.16} \times {(81)}^{0.09} \\ \\ \rm \: we \: know \: that \: {a}^{m} \rm \times {a}^{n} = {a}^{m + n} \\ \\ \rm therefore \: {(81)}^{0.16} \times {(81)}^{0.09} \\ \rm = {81}^{0.16 + 0.09} \\ \rm = {81}^{0.25} \\ \rm = {(3)}^{4×\frac{1}{4}} \\ \rm = \boxed{ 3} \: final \: answer \\ \\ \rm \: 2. \: given \: (2 + 3x)(2 - 3x) \\ \\ \rm by \: using \: {3}^{rd} law \: of \: algebraic \\ \rm expression \: that \: is \: (a - b) (a + b) \\ \rm = {a}^{2} - {b}^{2} \\ \\ \rm here \: a = 2 \: and \: b = 3x \\ \\ \rm therefore \: (2 + 3x)(2 - 3x) \\ \rm = {(2})^{2} - {(3x)}^{2} \\ \rm = \boxed{4 - 9 {x}^{2} } \: final \: answer \\ \\ \rm \: 3. given : - \\ \\ > > \rm \angle{A} = {40} \degree \\ > > \tt \angle{B} = {70} \degree \\ > > \rm \angle{C} = ? \\ \\ \rm \angle{A} + \angle{B} + \angle{C} = {180} \degree (angle \: \\ \rm sum \: property) \\ \rm {40} \degree + {70} \degree + \angle{C} = {180} \degree \\ \rm {110} \degree \: + \angle{C} = {180} \degree \\ \rm \angle{C} = {180} \degree - {110} \degree \\ \rm \angle{C} = {70} \degree \\ \\ \rm therefore \: \angle{DCE} = {70} \degree \: (vertical \\ \rm opposite \: angles) \\ \\ \rm and \: \angle{BCE} = {180} \degree - {70} \degree \\ \rm = \boxed{ {110} \degree} (linear \: pair)$

$\bf \large \it{Cheers!!!}$