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3 years ago

Ok so I barley downloaded this right now and it's a new update, how do u follow someone on here?

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

6 0

Yea i dont think you can follow any one on this app

olasank [31]3 years ago

4 0

I don’t think you can follow anyone on here tbh :/

You might be interested in

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

3 years ago

A kayaker moves 32 meters northward, then 6 meters

netineya [11]

Answer:

distance moved by kayaker is 54m

4 0

3 years ago

F(x) = 1/x g(x) = x - 4 Can you evaluate (gof)(0)? Explain why or why not.

s344n2d4d5 [400]

(gof)(0) cannot be evaluated

Solution:

Given that,

$f(x) = \frac{1}{x}\\\\g(x) = x - 4$

A composite function is denoted by (g o f) (x) = g (f(x)).

The notation g o f is read as “g of f”

Therefore, let us find whether (gof)(0) can be evaluated or not

To find (gof)(0):

(g o f) (x) = g (f(x))

Now substitute the given value of f(x)

$(g o f) (x) = g(\frac{1}{x})$

$\text{ Substitute } x = \frac{1}{x} \text{ in } g(x) = x - 4$

$(g o f) (x) = \frac{1}{x} - 4$

Now to find (gof)(0), substitute x = 0

$(g o f) (x) = \frac{1}{0} - 4$

Since 1 divided by 0 is undefined, because any number divided by 0 is undefined

(gof)(0) cannot be evaluated

6 0

3 years ago

Read 2 more answers

Please help me ASAP

Serggg [28]

The class 4S because you divide each number by its mean mark and standard division and 4S is the highest.

5 0

3 years ago

Help please will mark brainliest!!!

solniwko [45]

Answer:

945

Step-by-step explanation:

8 0

3 years ago