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3 years ago

Please answer the question .

Mathematics

1 answer:

Vsevolod [243]3 years ago

3 0

Answer:

The answer is A.

Step-by-step explanation:

I'm very sorry if this is wrong but A would make the most sense. the formula y=mx+b. m is the slope, and b is the y-intercept so A is the most reasonable.

If this helped you I would appreciate it if you marked me Brainliest. Thank you and have a nice day!

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${\large{\textsf{\textbf{\underline{\underline{Question \: 1 :}}}}}}$

$\star\:{\underline{\underline{\sf{\purple{Solution:}}}}}$

$\bullet \sf \: {(a + b)}^{ab}$

Putting value of a as 3 and b as -2, we get :

$\longrightarrow \sf \: {( 3 + (- 2))}^{3 \times - 2}$

$\longrightarrow \sf \: {( 3 - 2)}^{3 \times - 2}$

$\longrightarrow \sf \: {( 1)}^{ - 6}$

• Using negative Exponents Law

$\longrightarrow \sf \dfrac{1}{ {1}^{6} }$

$\longrightarrow \sf \dfrac{1}{ 1 \times 1 \times 1 \times 1 \times 1 \times 1 }$

$\longrightarrow \sf \dfrac{1}{ 1 }$

$\longrightarrow \sf \purple{1}$

${\large{\textsf{\textbf{\underline{\underline{Question \: 2 :}}}}}}$

$\star\:{\underline{\underline{\sf{\red{Solution:}}}}}$

$\bullet \sf \: \dfrac{ {8}^{ - 1} \times {5}^{3} }{ {2}^{ - 4}}$

$\longrightarrow \sf \: {8}^{ - 1} \times {5}^{3} \times \dfrac{1}{{2}^{ - 4}}$

• Using negative Exponents Law

$\longrightarrow \sf \: {8}^{ - 1} \times {5}^{3} \times {2}^{4}$

$\longrightarrow \sf \: {8}^{ - 1} \times 5 \times 5 \times 5 \times {2}^{4}$

$\longrightarrow \sf \: {8}^{ - 1} \times 125 \times {2}^{4}$

$\longrightarrow \sf \: {8}^{ - 1} \times 125 \times 2 \times 2 \times 2 \times 2$

• Using negative Exponents Law

$\longrightarrow \sf \: \dfrac{1}{ \cancel{8}_{4}} \times 125 \times \cancel{2}_{1} \times 2 \times 2 \times 2$

$\longrightarrow \sf \: \dfrac{1}{ \cancel4_{2}} \times 125 \times \cancel{2}_{1} \times 2 \times 2$

$\longrightarrow \sf \: \dfrac{1}{ \cancel2} \times 125 \times \cancel{2} \times 2$

$\longrightarrow \sf \: 125 \times 2$

$\longrightarrow \sf \red{ 250}$

${\large{\textsf{\textbf{\underline{\underline{Question \: 3 :}}}}}}$

$\star\:{\underline{\underline{\sf{\green{Solution(1):}}}}}$

$\bullet \sf \dfrac{ \sqrt{32} + \sqrt{48} }{ \sqrt{8} + \sqrt{12} }$

$\longrightarrow \sf \dfrac{ \sqrt{4 \times 4 \times 2} + \sqrt{4 \times 4 \times 3} }{ \sqrt{2 \times 2 \times 2} + \sqrt{2 \times 2 \times 3} }$

$\longrightarrow \sf \dfrac{ \sqrt{ {4}^{2} \times 2} + \sqrt{ {4}^{2} \times 3} }{ \sqrt{ {2}^{2} \times 2} + \sqrt{ {2}^{2} \times 3} }$

$\longrightarrow \sf \dfrac{ 4\sqrt{ 2} + 4 \sqrt{ 3} }{ 2\sqrt{ 2} +2 \sqrt{ 3} }$

$\longrightarrow \sf \dfrac{ \cancel{ 4}_{2}(\sqrt{ 2} + \sqrt{ 3}) }{ \cancel{2}(\sqrt{ 2} + \sqrt{ 3}) }$

$\longrightarrow \sf \dfrac{ 2 \: \cancel{(\sqrt{ 2} + \sqrt{ 3}) } }{ \cancel{(\sqrt{ 2} + \sqrt{ 3})} }$

$\longrightarrow \sf \green{2}$

$\star\:{\underline{\underline{\sf{\blue{Solution(2):}}}}}$

$\bullet \sf \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{80} + \sqrt{48} - \sqrt{45} - \sqrt{27} }$

$\begin{gathered} \longrightarrow \sf \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{4 \times 4 \times 5} + \sqrt{4 \times 4 \times 3} - \sqrt{3 \times 3 \times 5} - \sqrt{3 \times 3 \times 3} } \end{gathered}$

$\begin{gathered}\longrightarrow \sf \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{ {4}^{2} \times 5} + \sqrt{ {4}^{2} \times 3} - \sqrt{ {3}^{2} \times 5} - \sqrt{ {3}^{2} \times 3} } \end{gathered}$