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nataly862011 [7]

3 years ago

5

a ball is dropped from a height of 512 inches onto a level floor. after the fourth bounce it is still 2 inches off the ground. p

resuming that the height the ball bounces is always the same fraction of the height reached on the previous bounce. what is that fraction? A) 1/4 B) 3/7 C) 5/9 D)4/7 E)3/5

2 answers:

Stels [109]3 years ago

5 0

Answer:

A. 1/4

Step-by-step explanation:

We know that before the 1st bounce, the height of the ball is 512 inches.

Say the fraction is x.

Then, after the first bounce, the height of the ball is 512 * x = 512x.

After the second bounce, the height is now x * 512x = 512x².

By similar reasoning, the height after the third bounce is 512x³ and after the fourth bounce, it is $512x^4$ .

We also know that after the fourth bounce, the height is 2 inches. So, set 2 equal to $512x^4$ :

2 = $512x^4$

Divide both sides by 512:

$x^4=2/512$ $x^4=2/512=1/256$

Take the fourth root of both sides:

$x=\sqrt[4]{1/256} =1/4$

Hence, the answer is A.

<em>~ an aesthetics lover</em>

Sliva [168]3 years ago

5 0

Answer:

A. 1/4.

Step-by-step explanation:

This is exponential decay so we have , where x = the fraction:

512(x)^4 = 2

x^4 = 1/256

x= 1 / (256)^0.25

= 1/4

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alisha [4.7K]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

N/45=1/15<br> a. n= 1/3<br> b. n=4<br> c. n=675<br> d. n=3

jek_recluse [69]

Answer:

n=3

Step-by-step explanation:

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5 0

3 years ago

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Alla [95]

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Step-by-step explanation:

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3 years ago

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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

2 years ago