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Setler79 [48]
2 years ago
15

-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:

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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.

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Solve for t. You must write your answer in fully simplified form.<br> -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 <span><span><span>cos<span>(x)</span></span>=<span>Adjacent/Hypotenuse
</span></span></span>The inside trig function is <span><span>arccos<span>(<span>3/5</span>)</span></span></span>, which means <span><span><span>cos<span>(x)</span></span>=<span>3/5</span></span></span>. Comparing <span><span><span>cos<span>(x)</span></span>=<span>Adjacent/Hypotenuse</span></span></span> with <span><span><span>cos<span>(x)</span></span>=<span>3/5
</span></span></span>
Find <span><span>Adjacent=3</span></span> and <span><span>Hypotenuse=5.
</span></span>Then, using the Pythagorean theorem, find <span><span>Opposite=?
</span></span>a² = c² - b²
a² = 5² - 3² = 25 - 9 = 16
a = √16 = 4

<span><span>Adjacent=3</span></span><span><span>Opposite=4</span></span><span><span>Hypotenuse=5
</span></span><span>
Plug in the value for sin(x) = opposite/hypotenuse

sin(x) = 4/5 </span>
6 0
3 years ago
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