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AnnZ [28]
3 years ago
5

PA-ANSWER PO KAILANGAN NA PO

Mathematics
1 answer:
otez555 [7]3 years ago
8 0

16,20

5/6,6/7

4g(plus)4h,5i(plus)5j

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Brrunno [24]
The graph of -x + 3y = 6 will be a straight line passing through points (-6, 0) and (0, 2).
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Are the functions linearly dependent on the interval (0,1)? y1=sintcost,y2=4sin2ty1=sin⁡tcos⁡t,y2=4sin⁡2t hint: recall that func
VikaD [51]

Answer: hello your question is poorly written attached below is the complete question

answer:

since y1 = 1/2 y2 on ( 0,1 ) the function are linearly dependent on ( 0,1 ) ( B )

Step-by-step explanation:

First step : Determine if y1 and y2  are linearly dependent

interval = ( 0,1 )

Functions: y1 = sin t cost t ,   y2 = 4 sin t

We can verify if these functions are linearly independent or linearly dependent by using the determinate form

attached below is the detailed solution

From the attached solution we can conclude that

y1, y2 is linearly dependent on ( 0,1 )

hence ; since y1 = 1/2 y2 on ( 0,1 ) the function are linearly dependent on ( 0,1 )

4 0
3 years ago
Help algebra 2 please
34kurt

Answer:

option 4

Step-by-step explanation:

(f*g)(x) =(x² + x+ 1)*(x² - x -1)

           = x²*(x² - x -1)   + x(x² - x -1) + 1*( x² - x -1)

            = x²*x² - x²*x -x²*1   + x*x² - x*x -x*1 + x² - x -1

            = x⁴ - x³ - x² +x³ - x²  - x  + x² - x -1

           = x⁴  - x³ + x³  - x² - x² + x² - x - x - 1

       = x⁴ - x² - 2x - 1

4 0
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20 divided by one fourth
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\frac { 20 }{ \frac { 1 }{ 4 }  } \\ \\ =\frac { 20 }{ 1 } \cdot { \left( \frac { 1 }{ 4 }  \right)  }^{ -1 }

\\ \\ =\frac { 20 }{ 1 } \cdot { \left( \frac { 4 }{ 1 }  \right)  }^{ 1 }\\ \\ =\frac { 20 }{ 1 } \cdot \frac { 4 }{ 1 } \\ \\ =\frac { 80 }{ 1 } \\ \\ =80
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