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3 years ago

What is the domain of the function y=3 in x graphed below?

Mathematics

2 answers:

Semenov [28]3 years ago

4 0

Answer:

domain is x>0

Step-by-step explanation:

$y=3ln(x)$

domain is the set of x values for which the function is defined

ln(negative value ) is undefined

So we take positive values for x because we have ln(x)

From the graph we can see that the given equation has a graph for x values greater than 0

the graph goes close to 0 but does not crosses y axis

So domain is x>0

Jet001 [13]3 years ago

3 0

What is the domain of the function y=3 in x graphed below?

Domain is x>0, ln (0) is undefined and there is no negative ln.

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omeli [17]

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4 years ago

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FrozenT [24]

This is asking which one has the most factors. Lets list them out:

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3 years ago

Which of the following sets of data will have the smallest Standard Deviation?

PIT_PIT [208]

The formula for standard deviation is in attachment the mean (or average is subtracted from each data sample then squared, then summed up for all data samples, then divided by the number of samples, then finally the square root is taken.

The first row of data will have a standard deviation of 0.

The second row of numbers will have a low standard deviation because the differences between the sample data values are low ( <10).

The third row of number will have a somewhat larger standard because of the spread of the numbers.

The last row will have a slightly smaller standard deviation that the 3rd row because of the zero's

What smaller number can you get other than 0?

The first row of data has a 0 or smallest standard deviation

Download pdf

6 0

3 years ago

A= 12b1+b2h ; solve for b1

HACTEHA [7]

Answer:

b1 = 2A/h – b2

Step-by-step explanation:

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3 years ago

write an equation of the line that passes through (-1, 4) and is perpendicular to the line y=-1/3x+3

gladu [14]

Answer:

y= 3x + 7 Step-by-step explanation:

y= -1/3x+3 (so, gradient of this line is -1/3).

formula: gradient of line one x gradient of line two= -1 (applies for 2 perpendicular lines only).

So, m1 x -1/3= -1

m1= ( $\frac{-1}{\frac{-1} {3} }$ )

m1= 3 (gradient of perp line)

y=mx+c

4= 3 x -1 + c

7=c

therefore, y=3x+7 (eqn of the line perp to y=-1/3x+3)

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3 years ago