Can you help me with this? (I will give one of you brainliest)

Mathematics

2 answers:

Degger [83]3 years ago

8 0

Answer:

1) $\sqrt{9}$ and $\sqrt{16}$ (between 3 and 4)

Explanation: $\sqrt{9}$ and $\sqrt{16}$ are the closest WHOLE squares root to $\sqrt{12}$ ; any other square roots wouldn't give us a whole number, or they're farther away from

2) $-\sqrt{144}$ and $-\sqrt{169}$ (between -(12) and -(13))

3) $-\sqrt{36}$ and $-\sqrt{49}$ (between -(6) and -(7))

Juli2301 [7.4K]3 years ago

7 0

Answer:

see explanation

Step-by-step explanation:

Consider perfect squares on either side of the radicand

(a)

9 < 12 < 16 , then

$\sqrt{9}$ < $\sqrt{12}$ < $\sqrt{16}$ , so

3 < $\sqrt{12}$ < 4

(b)

- 144 < - $\sqrt{158}$ < - 169

- $\sqrt{144}$ < - $\sqrt{158}$ < - $\sqrt{169}$ , s0

- 12 < - $\sqrt{158}$ < - 13

(c)

- 36 < - $\sqrt{40}$ < - $\sqrt{49}$ , so

- 6 < - $\sqrt{40}$ < - 7

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Aleksandr-060686 [28]

Answer:

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${ \boxed{ \bf{ \frac{dy}{dx} = \frac{ {\huge{v} } \frac{du}{dx} - { \huge{u}} \frac{dv}{dx} }{ {v}^{2} } }}} \\$

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${ \rm{ \frac{dy}{dx} = \frac{(4 {x}^{2})( \frac{3}{2 \sqrt{x} }) - (3 \sqrt{x} + 2x)(8x) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{(6 {x}^{2})( {x}^{ - \frac{1}{2} } ) - (24 {x}^{ \frac{3}{2} } + 16 {x}^{2} ) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ 6 {x}^{ \frac{3}{2} } - {24x}^{ \frac{3}{2} } - {16x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ - 18 {x}^{ \frac{3}{2} } - 16 {x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = - 2 {x}^{ \frac{3}{2} } (9 + 8 {x}^{ \frac{4}{3} } ) \div 16 {x}^{4} }} \\ \\ { \boxed{ \boxed{ \rm{ \: \: \frac{dy}{dx} = - \frac{8 \sqrt{ {x}^{ \frac{8}{3} }} + 9}{8x {}^{ \frac{8}{3} } } \: \: }}}}$