If r=20.5 and s=34.2 find S Round to the nearest tenth

Mathematics

2 answers:

astraxan [27]4 years ago

7 0

Answer: option d

Step-by-step explanation:

Remember the identity:

$tan\alpha=\frac{opposite}{adjacent}$

The inverse of the tangent function is arctangent. You need to use this to calculate the angle "S":

$\alpha =arctan(\frac{opposite}{adjacent})$

Knowing that: you need to find the measure of the angle "S" , $r=20.5$ (which is the adjacent side) and $s=34.2$ (which is the opposite side), you can sustitute values into $\alpha =arctan(\frac{opposite}{adjacent})$

Then, you get that the measure of "S" rounded to the nearest tenth is:

$S=arctan(\frac{34.2}{20.5})\\\\S=59.1\°$

Arturiano [62]4 years ago

4 0

Answer:

59.1 °

Step-by-step explanation:

tan S = 34.2 / 20.5

S = tan -1 (34.2 / 20.5)

S = 59.1 ° Rounded to the nearest tenth!

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Y''-2y'+y=2(e^x)-3(e^-x)

labwork [276]

First find the characteristic solution. The characteristic equation is

$r^2-2r+1=(r-1)^2=0$

which as one root at $r=1$ of multiplicity 2. This means the characteristic solution for this ODE is

$y_c=C_1e^x+C_2xe^x$

For the nonhomogeneous part, you can try a particular solution of the form

$y_p=(a_2x^2+a_1x+a_0)e^x+be^{-x}$

which has derivatives

${y_p}'=(a_2x^2+(2a_2+a_1)x+a_1+a_0)e^x-be^{-x}$
${y_p}''=(a_2x^2+(4a_2+a_1)x+2a_2+2a_1+a_0)e^x+be^{-x}$

Substituting into the ODE, the left hand side reduces significantly to

$2a_2e^x+4be^{-x}=2e^x-3e^{-x}$

and it follows that

$\begin{cases}2a_2=2\\4b=-3\end{cases}\implies a_2=1,b=-\dfrac34$

Therefore the particular solution is

$y_p=x^2e^x-\dfrac34e^{-x}$

and so the general solution is the sum of the characteristic and particular solutions,

$y=y_c+y_p$
$y=C_1e^x+C_2xe^x+x^2e^x-\dfrac34e^{-x}$