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3 years ago

I need some help with this calculus 1 implicit diffrention problem?

Mathematics

1 answer:

katen-ka-za [31]3 years ago

7 0

Answer:

y" = -24 / y³

Step-by-step explanation:

6x² + y² = 4

Take the derivative of both sides with respect to x.

12x + 2y y' = 0

Again, take the derivative of both sides with respect to x.

12 + 2y y" + y' (2y') = 0

12 + 2y y" + 2(y')² = 0

Solve for y' in the first equation.

2y y' = -12x

y' = -6x/y

Substitute and solve for y":

12 + 2y y" + 2(-6x/y)² = 0

12 + 2y y" + 2(36x²/y²) = 0

12 + 2y y" + 72x²/y² = 0

6y² + y³ y" + 36x² = 0

y³ y" = -36x² − 6y²

y" = (-36x² − 6y²) / y³

Solve for y² in the original equation and substitute:

y² = 4 − 6x²

y" = (-36x² − 6(4 − 6x²)) / y³

y" = (-36x² − 24 + 36x²) / y³

y" = -24 / y³

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So the general expression for slices of pizza did Harris eat is

$\frac{n}{4}-1$

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Solve 7x − 2(x + 1) = 6x + 14. (5 points) −12 12 −16 16

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Step-by-step explanation:

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Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

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