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2 years ago

-10x + 10x + 5x + 5 = 5 solve for x

Mathematics

2 answers:

Maksim231197 [3]2 years ago

6 0

Answer:

x = 0

Step-by-step explanation:

n200080 [17]2 years ago

3 0

Answer:

x = 0

Step-by-step explanation:

-10x + 10x + 5x + 5 = 5

5x + 5= 5

5x = 0

x = 0

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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.

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

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2 years ago

Use the given data to find the equation of the regression line. examine the scatterplot and identify a characteristic of the dat

kari74 [83]

Answer:

y = 1.5x + 3.2009

Step-by-step explanation:

I plotted your points and drew a linear regression line.

Its equation is

y = 1.5x + 3.2009

The regression line ignores the fact that the points fall along a smooth curve.

A parabolic regression line fits the data much better.

The equation for that line is

y = - 0.1647x² + 4.1357x - 5.6946

It is the equation for a vertical parabola opening downward.

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3 years ago

Solve for t. You must write your answer in fully simplified form. -19 = 7t

a_sh-v [17]

Answer:

-2.71

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

3 1183 long division

Triss [41]

= 394 R 1

= 394 1/3

1183 divided by 3 equals
394 with a remainder of 1

3 0

3 years ago

Find the exact value of sin (arccos (3/5)). For full credit, explain your reasoning.

zloy xaker [14]

The equation for cosine is cos(x)=Adjacent/Hypotenuse
The inside trig function is arccos(3/5), which means cos(x)=3/5. Comparing cos(x)=Adjacent/Hypotenuse with cos(x)=3/5

Find Adjacent=3 and Hypotenuse=5.
Then, using the Pythagorean theorem, find Opposite=?
a² = c² - b²
a² = 5² - 3² = 25 - 9 = 16
a = √16 = 4

Adjacent=3Opposite=4Hypotenuse=5

Plug in the value for sin(x) = opposite/hypotenuse

sin(x) = 4/5

6 0

3 years ago