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3 years ago

Find the length of the third side. If necessary, write in simplest radical form

Mathematics

1 answer:

katrin [286]3 years ago

3 0

HERE,

the $\huge{\triangle}$ is a right angle traingle.

length of one side= $\bold{ 1 }$

length of another side = $\bold{\sqrt{3} }$

so,

the length of third side=

$\bold{\sqrt{(1)^{2} +({\sqrt{3}})^{2}} }$

$\bold{\sqrt{1+3} }$

$\bold{\sqrt{4} }$

${\huge{\fcolorbox{aqua}{navy}{\fcolorbox{yellow}{blue}{\bf{\color{yellow}{2}}}}}}$

$\therefore$ ,the length third side of the traingle is 2.

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Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term f here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

4 years ago

The ratio of girls to boys in Mr. Smith's homeroom is 3:4. The ratio in Mr. Smith's homeroom to students in the whole 7th grade

Leno4ka [110]

3:4 = G:B

1:5 = Mr. Smith's class: 7th Grade

2:7= 7th: Middle school

12 girls = 3 units

1 unit = 12/3= 4

Boys = 4x4= 16

Whole class = 28 students

Class : Grade = 1:5 7 = number of units in Mr. Smith's class

28 = 1 unit

5 units= 28x5= 140 There are 140 kids in the grade

140 = 2 units

1 unit = 140/2= 70

70x7=490

There are 490 students in the whole grade

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3 years ago

A library shelf has 3 3 types of books

AfilCa [17]

Answer:

that library has only 3 books? it needs to open a book raising fund

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3 years ago

I need help plss!! I don’t understand

MA_775_DIABLO [31]

Since it says what the area of the circle is, you need to work the problem backwards from finding the area. So it would be 100 ÷ 3.14 = 31.84 that is the radius. Next you need to find the diameter, so 31.84 × 2 = 63.68. Now you multiply 63.68 × 3.14 = 199.95, round that to 200, divide 200 ÷ 2 = 100 ÷ 2 = 50 miles

Answer= 50 miles

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Alexxandr [17]

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