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3 years ago

What figure has three sides that measure 2 cm, 5 cm, and 3 cm.

Mathematics

2 answers:

Rus_ich [418]3 years ago

7 0

Answer:

A quadrilateral shape has 2cm 5cm and 3 cm

Olegator [25]3 years ago

5 0

Answer:

None

Step-by-step explanation:

A three-sided polygon is a triangle.

In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side.

2 cm + 3 cm = 5 cm

5 cm = 5 cm

Since the sum of the lengths 2 cm and 3 cm is not greater than 5 cm, these three segment lengths cannot form a triangle.

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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.

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

3 0

3 years ago

Please help!!

Marysya12 [62]

The LCD of 3 and 4 is 12. To make 4 into 12, we multiply it by three, so 3 would also have to be multiply by 3, turning 3/4 into 9/12.

The same thing applies to 1/3, only it needs to be multiplied by 4. So it would become 4/12

9/12 - 4/12 = 5/12

4 0

3 years ago

Read 2 more answers

What needs to be done to 21/10 to complete the problem? If you can, try to fix it yourself.<br><br>

Olin [163]

Answer:

This is an improper fraction so we need to make it a mix fraction.

Step-by-step explanation:

21/10 = 10/10+10/10+1/10

10/10=1

So your final answer will be $2\frac{1}{10}$

8 0

3 years ago

Read 2 more answers

Write a system of linear equations for the graph below.

Vanyuwa [196]

Answer:

-3/-2

Step-by-step explanation:

because (x,y) the x coordinate is -3 and the y coordinate is -2

3 0

3 years ago

5:<br><br> will give brainliest

Fiesta28 [93]

The focal length of the given ellipse is given as (±6, 0)

<h3>Equation of an ellipse</h3>

An ellipse is defined as a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant or when a cone is cut by an oblique plane which does not intersect the base.

The standard equation of an ellipse is expressed as;

x^2/a^2 + y^2/b^2 = 1

The formula for calculating the focus of the ellipse is given as:

c^2 = b^2 - a^2

Given the equation of an ellipse

(x-7)^2/64 + (y-5)^2/100 = 1

This can also be expressed as:

(x-7)^2/8^2 + (y-5)^2/10^2 = 1

Comparing with the general equation

a = 8 and b = 10

Substitute

c^2 = 10^2 - 8^2

c^2 = 100 - 64

c^2 = 36

c = 6

Hence the focal length of the given ellipse is given as (±6, 0)

Learn more on focus of ellipse here; brainly.com/question/4429071

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6 0

2 years ago