Help me pls its Pythagorean theorem I need help

Mathematics

1 answer:

ankoles [38]3 years ago

8 0

Answer:

the answer is 3.81

but rounding it it will be 3.8 meters

Step-by-step explanation:

Pythagorean theorem says

a²+b²=c²

so,

take take the Lowest measurement as a² which will be 7 and take 8 as the hypotenuse or the longest side which is c²

then put it all together you get

7²+x²=8²

we made b as x because we will be finding b

then substitute the values

x²=8²-7²

x²=64-49

x²=15

to remove that square from the x you then put square root in both

✓x²=✓15

press in the calculator, root of 15 will be about 3.81

now we got

x=3.81

x is the missing measurement

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Ronch [10]

Answer:

X=7

Step-by-step explanation:

7*15=105 7-2=5 5*25=125 125+105=230

6 0

2 years ago

Use series to verify that <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.