Select the choice containing all of the congruent sides from the two triangles.

Mathematics

1 answer:

Lynna [10]3 years ago

5 0

Answer:

Step-by-step explanation:

The order of the letters in the side's name should correspond to their position in the name of the triangle.

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Help pleaseeeee!!!!!!

never [62]

The square it’s self without the side triangles is 132ft^2 and the triangles them selves are 18ft^2.
Since there are 2 triangles both combined would be a total of 36ft^2
The total area of the square is 132ft^2
Total area of one triangle is 18ft^2
Total area is 168ft^2

Side length dimensions:
Square without sides: 12x11 ft
Each triangle: 1/2 12x3

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3 years ago

A ladder leaning against a wall forms a triangle and exterior angles with the wall and the ground. What are the measures of the

Paha777 [63]

11x° is interior angle of ladder to the ground
and 7X° is interior angle of ladder to the wall

exterior angle of a ladder to wall is 180-7x
and exterior angle of a ladder to ground is 180-11x

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3 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.