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Dmitry_Shevchenko [17]
3 years ago
9

PLEASEEE HELP!!!!

Mathematics
2 answers:
notsponge [240]3 years ago
5 0
Can you space them out I cant see em all if there are together
jarptica [38.1K]3 years ago
5 0

Answer:

f(x)=4-4x ;

Domain:{0,1,3,5,6}

Range;{-20,-16,-8,0,4}

f(x)=5x-3

Domain:{-2,-1,0,3,4}

Range:{-13,-8,-3,12,4}

f(x)=-10x

Domain:{-4,-2,0,2,4}

Range:{-40,-20,0,20,40}

f(x)=\frac{3}{x}+1.5

Domain:{-3,-2,-1,2,6}

Range:{0.5,0,-1.5,3,2}.

Step-by-step explanation:

<h3>First  we take function f(x)=4-4x</h3>

Take Domain:{0,1,3,5,6}

If, we take x=0 and put in the function then we get

f(x)=4-4\times0

f(x)=4-0

f(x)=4

put x=1

then f(x)=4-4\times1

f(x)=4-4=0

put x=3 then we get

f(x)=4-12=--8

put x=5 them we get

f(x)=4-20=-16

put x=6 then we get

f(x)=4-24=-20

Therefore ,range:[-20,-16,-8,0,4}

<h3>Now ,we take function f(x)=5x-3</h3>

Take domain{-2,-1,0,3,4}

Now, put x=-2 in the function then we get

f(x)=5\times(-2)-3

f(x)=-13

now put x=-1 then we get

f(x)=-5-3=-8

Put x=0 then we get

f(x)=0-3=-3

Put x=3 then we get

f(x)=15-3=12

Put x=4 then we get

f(x)=20-3=17

Therefore , range:{-13,-8,-3,12,17}

Now, we take III function f(x)=-10x

Take domain:{-4,-2,0,2,4}

Put x=-4 in the function then we get

f(x)=10\times[tex]10\times(-4)

f(x)=-40

Put x= -2 then we get

f(x)=10\times(-2)=-20

Put x=0 then we get

f(x)=0

Put x=2 then we get

f(x)=20

Put x=4 then we get

f(x)=40

Therefore , range :{-40,-20,0,20,40}

Now, we take IV function f(x)=\frac{3}{x}+1.5

Take domain:{-3,-2,-1,2,6}

Put x= -3 in the taken function then we get

f(x)=-1+1.5=0.5

put x=-2 then  we get

f(x)= -1.5+1.5=0

Put x=-1 then we get

f(x)=-3+1.5=-1.5

Put x= 2 then we get

f(x)=1.5+1.5=3

Put x= 6 then we get

f(x)=0.5+1.5=2

Therefore, range : {0.5,0,-1.5,3,2}.

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Nuetrik [128]

Answer:

y(s) = \frac{5s-53}{s^{2} - 10s  + 26}

we will compare the denominator to the form (s-a)^{2} +\beta ^{2}

s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}

comparing coefficients of terms in s

s^{2} : 1

s: -2a = -10

      a = -2/-10

      a = 1/5

constant: a^{2}+\beta ^{2} = 26

               (\frac{1}{5} )^{2} + \beta ^{2} = 26\\\\\beta^{2} = 26 - \frac{1}{10} \\\\\beta =\sqrt{26 - \frac{1}{10}} =5.09

hence the first answers are:

a = 1/5 = 0.2

β = 5.09

Given that y(s) = A(s-a)+B((s-a)^{2} +\beta ^{2} )

we insert the values of a and β

  \\5s-53 = A(s-0.2)+B((s-0.2)^{2} + 5.09^{2} )

to obtain the constants A and B we equate the numerators and we substituting s = 0.2 on both side to eliminate A

5(0.2)-53 = A(0.2-0.2) + B((0.2-0.2)²+5.09²)

-52 = B(26)

B = -52/26 = -2

to get A lets substitute s=0.4

5(0.4)-53 = A(0.4-0.2) + (-2)((0.4 - 0.2)²+5.09²)

-51 = 0.2A - 52.08

0.2A = -51 + 52.08

A = -1.08/0.2 = 5.4

<em>the constants are</em>

<em>a = 0.2</em>

<em>β = 5.09</em>

<em>A  = 5.4</em>

<em>B = -2</em>

<em></em>

Step-by-step explanation:

  1. since the denominator has a complex root we compare with the standard form s^{2} -10s+26 = (s-a)^{2} +\beta ^{2} = s^{2} -2as +a^{2} +\beta ^{2}
  2. Expand and compare coefficients to obtain the values of a and <em>β </em>as shown above
  3. substitute the values gotten into the function
  4. Now assume any value for 's' but the assumption should be guided to eliminate an unknown, just as we've use s=0.2 above to eliminate A
  5. after obtaining the first constant, substitute the value back into the function and obtain the second just as we've shown clearly above

Thanks...

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