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3 years ago

Help me for 19 points please thank you

Mathematics

1 answer:

valina [46]3 years ago

3 0

Answer:

Step-by-step explanation:

I hope you dont have to show your work because I am not showing how you do all of those aha.

part 1:

1. p=72, a= 612

2. p=194, a= 32.5

3. p=300, a= 1500

4. p=221, a= 35

and um... I dont know how to do part two. sorry

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Find an equation of a line that is tangent to the graph of f and parallel to the given line. Please see picture

ivanzaharov [21]

Answer:

y = 3x - 2 (smaller y-intercept)

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Step-by-step explanation:

First let's write the generic equation of a line:

y = ax + b

This line needs to be parallel to the line 3x - y + 5 = 0, so it needs to have the same slope of this line.

The line 3x - y + 5 = 0 has a slope of 3, so our line has a = 3:

y = 3x + b

Now we need to find the values of b that make this line tangent to the function f(x) = x^3

Let's first find the derivative of f(x) in relation to x:

df(x)/dx = 3x^2

This derivative is the slope of the tangent line to the function for any value of x. We need a slope of 3, so:

3x^2 = 3

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x = ±1

Now, to find the y-values, we have:

f(1) = 1^3 = 1

f(-1) = (-1)^3 = -1

So, using the points (1,1) and (-1,-1) in our parallel line, we have:

first line using (1,1) : 1 = 3*1 + b

b = -2

second line using (-1,-1) : -1 = -3*1 + b

b = 2

The value of b is the y-intercept of the line, so the line with smaller y-intercept is y = 3x - 2, and the line with larger y-intercept is y = 3x + 2

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Step-by-step explanation:

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