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3 years ago

HELPPPPP

Mathematics

2 answers:

soldi70 [24.7K]3 years ago

8 0

x=50

Step-by-step explanation:

angle 1 and 3 are corresponding angles that menas they equal the same

so set it up like this 2x+10=110 solve for x

move over the 10, subtract 10 from 110 you get 100 and divide by 2

you get 50.

tamaranim1 [39]3 years ago

7 0

<h3>Answer: A) 50</h3>

=================================================

Explanation:

Angles 1 and 3 are vertical angles which are congruent.

angle 1 = angle 3

110 = 2x+10

2x+10 = 110

2x = 110-10

2x = 100

x = 100/2

x = 50

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Hallar el valor de ab datos: bc= 4,5 cm de= 2cm ef= 3 cm

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The measure of ab and df are 3cm and 7.5cm respectively

<h3 /><h3>Similarity theorem of shapes</h3>

In order to determine the missing sides, we need to take the ratio of the similar sides as shown:

1) ab/de = ac/df

ab/de = ab+bc/df

Substitute the given values

ab/2 = ab+4.5/2+3

ab/2 = ab+4.5/5

2(ab+4.5) = 5ab
2ab+9=5ab

3ab = 9

ab = 3cm

Hence the measure of ab is 3cm

2) ae/ce = bf/df

24/9 = 20/df

24df =. 9(20)

24df = 180

df = 7.5cm

Hence the measure of ab and df are 3cm and 7.5cm respectively

Learn more on similar shapes here: brainly.com/question/11920446

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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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