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g100num [7]
3 years ago
8

HELPPPP WHICH 2 ARE LINEAR FUNCTIONS ???

Mathematics
2 answers:
DerKrebs [107]3 years ago
8 0

Answer:

1 and 2

Step-by-step explanation:

The first and second option,

The others all have a number or variable with exponent which means it's a parabola and linear means straight lines and parabolas are no straight at all.

emmasim [6.3K]3 years ago
3 0
The linear ones are the top 2
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Select TWO values of x that are roots
Sauron [17]

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Answer: A and B.

3 0
3 years ago
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There are 205 total students in a school. 20% of the students are 5th
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Answer:

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Step-by-step explanation:

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3 years ago
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Find the derivative of the function y = 2x^2 - 13x + 5 and use it to find the equation of the line tangeant to the curve at x =
ser-zykov [4K]
Y = 2x^2 - 13x + 5
dy/dx = 4x - 13

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The tangent line at x = 3, passes through the point (3, -16) and has a slope of -1
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3 0
3 years ago
Find the critical numbers of the function. (enter your answers as a comma-separated list. if an answer does not exist, enter dne
NNADVOKAT [17]

The critical values of the function are x = 0 and x = 64

The critical values of a function f(x) are the values of x for which f'(x) = 0.

Given,

f(x) = \frac{x^{4} }{5(x-8)^{2} }

The derivative is found as follows, applying the quotient rule:

f'(x) = \frac{[5(x^{4} )(x-8)^{2}-5x^{4} [(x-8)^{2} ]' }{[5(x-8)^{2} ]^{2}  }

     = \frac{2x^{3}(x-64) }{5(x-8)^{3} }

The zeros of the function are the zeros of the numerator, thus:

2x^{3} (x-64) =0\\2x^{3} =0-------- > x =0\\

x - 64 = 0------------->x = 64

The critical values are x = 0 and x = 64

Learn more about critical values here:brainly.com/question/12958088

#SPJ4

3 0
1 year ago
11.05+14.6+46<br> SHOW WORK plz
mart [117]

1  
11.05
14.60
46.00
71.65   is your answer
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3 years ago
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