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zlopas [31]
2 years ago
15

Which relation is NOT a function? Explain your answer.

Mathematics
1 answer:
maw [93]2 years ago
8 0
#2 or B is not a function because it has 3 as X twice
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The number of hours you work is dependent and so so
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3 years ago
What is the surface area of the following cylinder in square centimeters? 54.95 cm2 471 cm2 47.1 cm2 31.4 cm2
goldfiish [28.3K]

Answer:

54.95

hope this helps :p

5 0
3 years ago
Consider the initial value function y given by
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...

3 0
3 years ago
What is the domain of the function f(x) = x^2 + 5?
Sophie [7]

The domain is R so the first one is correct.

4 0
3 years ago
Can Someone help me. PLeeeease..............
aivan3 [116]
I believe the answer is 24 times
hope this helps!
7 0
3 years ago
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