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LiRa [457]
2 years ago
6

Can anyone help me out with this?​

Mathematics
1 answer:
bogdanovich [222]2 years ago
5 0

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

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

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

<u>Putting value of a as 3 and b as -2, we get</u><u> </u><u>:</u>

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

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

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

• <u>Using negative Exponents Law</u>

\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}}

<u>• Using negative Exponents Law</u>

\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

<u>• Using negative Exponents Law</u>

\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}

\longrightarrow \sf  \dfrac{ \sqrt{5}  +  \sqrt{3} }{4 \sqrt{  5} + 4 \sqrt{   3}  - 3\sqrt{    5}  -  3\sqrt{  3}   }

\longrightarrow \sf  \dfrac{ \sqrt{5}  +  \sqrt{3} }{4 \sqrt{  5}   - 3\sqrt{    5} + 4 \sqrt{   3} -  3\sqrt{  3}   }

\longrightarrow  \sf \dfrac{ \cancel{ \sqrt{5}  +  \sqrt{3}} }{ \cancel{\sqrt{    5}  +   \sqrt{  3}   } }

\longrightarrow   \blue{1}

{\large{\textsf{\textbf{\underline{\underline{Answers :}}}}}}

• Question 1 - \purple{1}

• Question 2 - \red{250}

• Question 3(1) - \green{2}

• Question 3(2) - \blue{1}

{\large{\textsf{\textbf{\underline{\underline{ Concept \: :}}}}}}

<u>★</u><u> </u><u>Negative</u><u> Exponents Law -</u>

\bullet  \sf \:  {a}^{ - m}  =  \dfrac{1}{ {a}^{m} }

★ \sqrt{32} can be written as 4 \sqrt{2}

‣ \sqrt{48} can be written as 4 \sqrt{3}

‣ \sqrt{8} can be written as 2 \sqrt{2}

‣ \sqrt{12} can be written as 2 \sqrt{3}

‣ \sqrt{80} can be written as 4 \sqrt{5}

‣ \sqrt{48} can be written as 4 \sqrt{3}

‣ \sqrt{45} can be written as 3 \sqrt{5}

‣ \sqrt{27} can be written as 3 \sqrt{3}

★ <u>During Addition and Subtraction</u>

• minus (-) minus (-) gives plus (+)

• minus (-) plus (+) gives minus (-)

• plus (+) minus (-) gives minus (-)

• plus (+) plus (+) gives plus (+)

• Also the sign of the resultant term depends upon the sign of the largest number.

{\large{\textsf{\textbf{\underline{\underline{ Note \: :}}}}}}

• Swipe to see the full answer.

\begin{gathered} {\underline{\rule{330pt}{3pt}}} \end{gathered}

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What does y equal in this equation 5(y+2/5)=-13
Korolek [52]

Steps to solve:

5(y + 2/5) = -13

~Distributive Property

(5 * y) + (5 * 2/5) = -13

~Simplify

5y + 2 = -13

~Subtract 2 to both sides

5y + 2 - 2 = -13 - 2

~Simplify

5y = -15

~Divide 5 to both sides

5y/5 = -15/5

~Simplify

y = -3

Best of Luck!

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HELP ME ASAP PLEASE<br> 1/5 X- 2/3 equals 4/3.<br> The solution is X equals what??
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Step-by-step explanation:

1/5 X- 2/3=4/3

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kotegsom [21]

Answer:

The distance between the points is approximately 6.4

Step-by-step explanation:

The given coordinates of the points are;

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The distance between two points, 'A', and 'B', on the coordinate plane given their coordinates, (x₁, y₁), and (x₂, y₂) can be found using following formula;

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Substituting the known 'x', and 'y', values for the coordinates of the points, we have;

l_{(2, \, -2), \ (6, \, 3) } = \sqrt{\left (3-(-2)  \right )^{2}+\left (6-2  \right )^{2}} = \sqrt{5^2 + 4^2} = \sqrt{41}

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3 years ago
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