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MAVERICK [17]
3 years ago
15

In an experiment, the one variable that is changed by the person doing the experiment is called the Question 2 options: controll

ed variable testing variable independent variable dependent variable
Mathematics
1 answer:
Ad libitum [116K]3 years ago
5 0

Answer: Independent Variable

Step-by-step explanation:

The independent (or manipulated) variable is something that the experimenter purposely changes or varies over the course of the investigation. The dependent (or responding) variable is the one that is observed and likely changes in response to the independent variable.

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Simplify 3(2m^3n^-2)^4
Effectus [21]

Answer:

(48 m^12)/n^8

Step-by-step explanation:

Simplify the following:

3 ((2 m^3)/(n^2))^4

Multiply each exponent in (2 m^3)/(n^2) by 4:

3×2^4 m^(4×3) n^(-2×4)

4 (-2) = -8:

3×2^4 m^(4×3) n^(-8)

4×3 = 12:

(3×2^4 m^12)/(n^8)

2^4 = (2^2)^2:

(3 (2^2)^2 m^12)/(n^8)

2^2 = 4:

(3×4^2 m^12)/(n^8)

4^2 = 16:

(3×16 m^12)/(n^8)

3×16 = 48:

Answer: (48 m^12)/n^8

5 0
3 years ago
Another truck with the same capacity travels from the city of Puebla to Tlaxcala. Only this truck carries passengers with 115% o
Juliette [100K]

Answer:d

Step-by-step explanation:

d

3 0
2 years ago
What is the problem of this solving?!
nika2105 [10]
     This question can be solved primarily by L'Hospital Rule and the Product Rule.

y= \lim_{x \to 0}  \frac{x^2cos(x)-sin^2(x)}{x^4}
 
     I) Product Rule and L'Hospital Rule:

y= \lim_{x \to 0} \frac{[2xcos(x)-x^2sin(x)]-2sin(x)cos(x)}{4x^3}
 
     II) Product Rule and L'Hospital Rule:

y= \lim_{x \to 0} \frac{[-2xsin(x)+2cos(x)]-[2xsin(x)+x^2cos(x)]-[2cos^2(x)-2sin^2(x)]}{12x^2} \\ y= \lim_{x \to 0} \frac{2cos(x)-4xsin(x)-x^2cos(x)-2cos^2(x)+2sin^2(x)}{12x^2}
 
     III) Product Rule and L'Hospital Rule:

]y= \alpha + \beta \\ \\ \alpha =\lim_{x \to 0} \frac{-2sin(x)-[4sin(x)+4xcos(x)]-[2xcos(x)-x^2sin(x)]}{24x} \\ \beta = \lim_{x \to 0} \frac{4sin(x)cos(x)+4sin(x)cos(x)}{24x} \\  \\ y = \lim_{x \to 0} \frac{-6sin(x)-4xcos(x)-2xcos(x)+x^2sin(x)+8sin(x)cos(x)}{24x}
 
     IV) Product Rule and L'Hospital Rule:

y = \phi + \varphi \\  \\ \phi = \lim_{x \to 0}  \frac{-6cos(x)-[-4xsin(x)+4cos(x)]-[2cos(x)-2xsin(x)]}{24x}  \\ \varphi = \lim_{x \to 0}  \frac{[2xsin(x)+x^2cos(x)]+[8cos^2(x)-8sin(x)]}{24x}
 
     V) Using the Definition of Limit:

y= \frac{-6*1-4*1-2*1+8*1^2}{24}  \\ y= \frac{-4}{24}  \\ \boxed {y= \frac{-1}{6} }
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3 years ago
Multiple choice question
Law Incorporation [45]

Answer:

B, C, and D

Step-by-step explanation:

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3 years ago
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