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mojhsa [17]
3 years ago
15

Tom's coach keeps track of the number of plays that Tom carries the ball and how many yards he gains. Select all the statements

about independent and dependent variables that are true.
Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
6 0

Answer:

The dependent variable is the number of plays he carries the ball.

The independent variable is the number of touchdowns he scores.

The dependent variable is the number of yards he gains.

Step-by-step explanation:

Tom's coach keeps track of the number of plays that Tom carries the ball and how many yards he gains. Select all the statements about independent and dependent variables that are true.

The dependent variable is the number of plays he carries the ball.

The independent variable is the number of plays he carries the ball.

The independent variable is the number of touchdowns he scores.

The dependent variable is the number of yards he gains.

The dependent variable is the number of touchdowns he scores.

y = kx

y = dependent variable. it is the variable that is explained by the dependent variable

x = is the variable that explains the dependent variable

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PLS HELP I NEED AN ANSWER
meriva

Answer:

B.

Step-by-step explanation:

None of the other answers make sense, so just by process of elimination it should be B.

Good luck on your assignment!

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3 years ago
Can anyone help me out with this?​
bogdanovich [222]

{\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}

5 0
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An outdoor staircase has a total vertical elevation of 17.5 feet from the road level to the first floor. The 1st floor to the 2n
docker41 [41]
See the attached picture for the answers:

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G(x) {-x if x&lt;-5. <br> 3 if x &gt;-5
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pakyu all

Step-by-step explanation:

ahvsjsvwkdsnnsjsbshsbajsbsjsvhs

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