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mr_godi [17]
3 years ago
13

What is the distance between the two endpoints in the graph below? If necessary, round your answer to two decimal places.

Mathematics
1 answer:
melisa1 [442]3 years ago
8 0
Answer: Choice D) 7.21 units

-----------------------------------------------------------------------------------------------

Work Shown:

Using the distance formula, we get
d = sqrt( (x2-x1)^2 + (y2-y1)^2 )
d = sqrt( (1-(-3))^2 + (-3-3)^2 )
d = sqrt( (1+4)^2 + (-3-3)^2 )
d = sqrt( (4)^2 + (-6)^2 )
d = sqrt( 16 + 36 )
d = sqrt( 52 )
d = 7.21110255 *** see note below ***
d = 7.21

Note: Use a calculator with a square root function. The value is approximate.
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(5.9)(1.102)<br><br><br> What’s does that mean
Leona [35]

Answer:

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Find the fraction of stundents enrolled in each language
blondinia [14]

Fraction of students enrolled in Chinese = \frac{51}{126}

Fraction of students enrolled in French = \frac{33}{126}

Fraction of students enrolled in Spanish = \frac{42}{126}

Solution:

Total number of students = 51 + 33 + 42

                                          = 126

Number of students enrolled in Chinese = 51

Fraction of students enrolled in Chinese

             $=\frac{\text{Number of students enrolled in Chinese}}{\text{Total number of students}}

             $=\frac{51}{126}

Fraction of students enrolled in Chinese = \frac{51}{126}

Number of students enrolled in French = 33

Fraction of students enrolled in French

             $=\frac{\text{Number of students enrolled in French}}{\text{Total number of students}}

             $=\frac{33}{126}

Fraction of students enrolled in French = \frac{33}{126}

Number of students enrolled in Spanish = 42

Fraction of students enrolled in Spanish

             $=\frac{\text{Number of students enrolled in Spanish}}{\text{Total number of students}}

             $=\frac{42}{126}

Fraction of students enrolled in Spanish = \frac{42}{126}

7 0
3 years ago
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Aleksandr-060686 [28]

Answer:

{ \rm{y =  \frac{3 \sqrt{x}  + 2x}{ {4x}^{2} } }} \\

• We are gonna use the quotient rule

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

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  • v is 4x²
  • du/dx is 3/2√x
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• Therefore:

{ \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} } }  \:  \: }}}}

5 0
3 years ago
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Can someone please please help me?? :(
salantis [7]

Answer:

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Step-by-step explanation:

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=6/8 (simplifies down to 3/4)

8 0
3 years ago
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