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3 years ago

Suppose x and y are related by the given equation and use implicit differentiation to determine dydx.x7y+y7x=7.

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

6 0

Looks like the equation is

$x^7y+y^7x=7$

Differentiate both sides with respect to $x$ , taking $y$ to be a function of $x$ .

$\dfrac{\mathrm d[x^7y+y^7x]}{\mathrm dx}=\dfrac{\mathrm d[7]}{\mathrm dx}$

$\dfrac{\mathrm d[x^7]}{\mathrm dx}y+x^7\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\mathrm d[y^7]}{\mathrm dx}x+y^7\dfrac{\mathrm dx}{\mathrm dx}=0$

$7x^6y+x^7\dfrac{\mathrm dy}{\mathrm dx}+7y^6x\dfrac{\mathrm dy}{\mathrm dx}+y^7=0$

Solve for $\frac{\mathrm dy}{\mathrm dx}$ :

$(x^7+7y^6)\dfrac{\mathrm dy}{\mathrm dx}=-(7x^6y+y^7)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{7x^6y+y^7}{x^7+7y^6}$

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"is it appropriate to use the normal approximation for the sampling distribution of"

Katyanochek1 [597]

Answer: Normal approximation can be used for discrete sampling distributions, such as Binomial distribution and Poisson distribution if certain conditions are met.

Step-by-step explanation: We will give conditions under which the Binomial and Poisson distribitions, which are discrete, can be approximated by the Normal distribution. This procedure is called normal approximation.

1. Binomial distribution: Let the sampling distribution be the binomial distribution $B(n,p)$ , where $n$ is the number of trials and $p$ is the probability of success. It can be approximated by the Normal distribution with the mean of $np$ and the variance of $np(1-p)$ , denoted by $N(np,np(1-p))$ if the following condition is met:

$n>9\left(\frac{1-p}{p}\right)\text{ and } n>9\left(\frac{p}{1-p}\right)$

2. Poisson distribution: Let the sampling distribution be the Poisson distribution $P(\lambda)$ where $\lambda$ is its mean. It can be approximated by the Normal distribution with the mean $\lambda$ and the variance $\lambda$ , denoted by $N(\lambda,\lambda)$ when $\lambda$ is large enough, say $\lambda>1000$ (however, different sources may give different lower value for $\lambda$ but the greater it is, the better the approximation).

5 0

3 years ago

What is a cube in math

DerKrebs [107]

Answer:

In arithmetic and algebra, the cube of a number n is its third power, that is, the result of multiplying three instances of n together. The cube is also the number multiplied by its square: n3 = n × n2 = n × n × n.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

An employee puts 10000 in a retirement account that offers 4% interest. How much will he have in 12 years

steposvetlana [31]

Answer:

14800

Step-by-step explanation:

The formula for simple interest (I) in terms of principal (P), rate (R) and time (T) is given as follows;

I = P x R x T / 100 ------------- (i)

Given:

Principal (P) = Initial amount being put into the account = 10000

Rate (R) = The interest rate being offered by the account manager = 4%

Time (T) = Time taken = 12 years

Substitute these values into equation (i) as follows:

I = 10000 x 4 x 12 / 100

I = 4800

Therefore, the initial amount will yield an interest of 4800 for those 12 years.

The total amount the employee will thus have in 12 years will be the sum of the initial amount and the interest. i.e

Amount = P + I

Amount = 10000 + 4800

Amount = 14800

4 0

3 years ago

Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.

elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

$\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}$

Consider a vector field $F=x\^i+y\^j+z\^k$

Then the divergence of the vector F is,

div F = $\nabla.F$ = $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)$

and the curl of the vector F is,

curl F = $\nabla\times F$ = $\^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})$

<h3>Calculation:</h3>

The given vector fields are:

$F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$

1) Verifying the identity: $\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

Consider L.H.S

⇒ $\nabla.(aF1+bF2)$

⇒ $\nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))$

⇒ $\nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the dot product between these two vectors,

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(1)

Consider R.H.S

⇒ $a\nabla.F1+b\nabla.F2$

So,

$\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)$

⇒ $\nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}$

$\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)$

⇒ $\nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}$

Then,

$a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})$

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(2)

From (1) and (2),

$\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

2) Verifying the identity: $\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Consider L.H.S

⇒ $\nabla\times(aF1+bF2)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the cross product,

$\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})$ ...(3)

Consider R.H.S,

⇒ $a\nabla\times F1+b\nabla\times F2$

So,

$a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))$

⇒ $\^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})$

$a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))$

⇒ $\^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})$

Then,

$a\nabla\times F1+b\nabla\times F2$ =

...(4)

Thus, from (3) and (4),

$\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Learn more about divergence and curl of a vector field here:

brainly.com/question/4608972

#SPJ4

Disclaimer: The given question on the portal is incomplete.

Question: Let $F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$ be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

$1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

8 0

1 year ago

Something on sale for $725. The original price was $1299. To the nearest tenth of a percent, what was the rate of discount?

ElenaW [278]

Answer:

44.2%

Step-by-step explanation:

Rate of discount = (difference in prices / original price) x 100

Difference in price = $1299 - $725 = $574

Original price = $1299

($574 / $1299) x 100 = 44.187837%

the tenth is the first number after the decimal place. To convert to the nearest tenth, look at the number after the tenth (the hundredth). If the number is greater or equal to 5, add 1 to the tenth figure. If this is not the case, add zero

the hundredth figure is 8, which is greater than 5. 1 would be added to the tenth figure.

So it becomes 44.2%

4 0

2 years ago