3/4x-2/3=5/12. i need help

The slope of the the tangent line to a curve at a point. Calc derivative help

MrRissso [65]

Hello from MrBillDoesMath!

Discussion:

The slope of the tangent to the curve y = 4x^3 is given by the first derivative.

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y' = 4 (3x^2) = 12x^2

At (-3,-108), the slope is 12x^2 = 12(-3)^2 = 12*9 = 108

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y = mx + b

From the first part, m = 108 so y = 108x + b.

Substituting y = -108 when x = -3 gives

-108 = 108(-3) + b => add 3* 108 to both sides

-108 + 108(3) = b => as 3 -1 = 2

108*2 = b = 216

y = 108x + 216

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Thank you,

MrB

5 0

3 years ago

TELL ME THIS ANSWER

V125BC [204]

Answer:

hope this helps you

have a great day God bless you

3 0

3 years ago

Read 2 more answers

Complete the equation of variation where y varies inversely as x and y = 80 when x = 0.7.

andre [41]

Y varies inversely with x
means that y=k/x
with x=0.7 and y=80
we have 80=k/0.7
k=0.7x80=56
so the equation is
y=56/x

3 0

3 years ago

Read 2 more answers

The number of motor vehicles registered in millions in the US has grown as follows:

zysi [14]

Answer:

a) Figure attached

b) $r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=10 $\sum x = 75.819, \sum y = 43.5231, \sum xy = 330.0321, \sum x^2 =574.8598, \sum y^2 =192.8274$

$r=\frac{10(330.0321)-(75.81948)(43.5231)}{\sqrt{[10(574.8598) -(75.819)^2][10(192.8274) -(43.5231)^2]}}=0.989$

So then the correlation coefficient would be r =0.989

Step-by-step explanation:

Previous concepts

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

Solution to the problem

Part a

Year (x): 1940, 1945 1950, 1955, 1960, 1965, 1970, 1975, 1980, 1985

Vehicles (Y): 32.4, 31.0, 49.2, 62.7, 73.9, 90.4, 108.4, 132.9, 155.8, 171.7

After apply natural log for the two variables and create the scatterplot in excel we got the following result on the figure attached.

Part b

And in order to calculate the correlation coefficient we can use this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=10 $\sum x = 75.819, \sum y = 43.5231, \sum xy = 330.0321, \sum x^2 =574.8598, \sum y^2 =192.8274$

$r=\frac{10(330.0321)-(75.81948)(43.5231)}{\sqrt{[10(574.8598) -(75.81948)^2][10(192.8274) -(43.5231)^2]}}=0.989$

So then the correlation coefficient would be r =0.989

4 0

3 years ago

Jolene wants to invest $2000 into a money market account. Interest is to be 6.5% compounded annually, and she wants to take the

timama [110]

Answer:

Interest = 650$

Step-by-step explanation:

Given:

Invest Inv = 2000$, interest rate r = 6.5% compounded annually

and time n = 5 years

Formula for calculation is:

Int = (Inv · r · n) / 100

Int = (2000 · 6.5 · 5) / 100 = 650$

Int = 650$

God with you!!!

3 0

3 years ago