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3 years ago

(Please answer quickly)

Mathematics

2 answers:

matrenka [14]3 years ago

5 0

Answer:8

Step-by-step explanation: Pythagorean theorem 3 4 5 multiply all by 2 3*2=6 4*2=8 5*2=10

Elenna [48]3 years ago

4 0

Answer:

huge erection

Step-by-step explanation:

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Help please hurry !!!!!!

forsale [732]

Answer:

Step-by-step explanation:

circumference= 29.83

2 times 3.14 times radius

area=70.83

3.14 times radius squared

3. radius

4. Diameter

radius is 4.75

diameter is 9.5

not 100% sure on finding the angle measure. looks about 40 degrees

8 0

3 years ago

Taryn bought 12.6 gallons of gasoline. There are approximately 3.8 liters in 1 gallon. Which measurement is closest to the numbe

Alina [70]

Answer:

<h2> Amount in liters is 47.9 liters</h2>

Step-by-step explanation:

The Question is incomplete, it does not provide the measurements or options to choose from, but we can estimate the answer

Step one:

Given data

Taryn Bought 12.6gallons of gasoline

1 gallon of gasoline is 3.8 liters

We want to convert from gallons to liters

Step two:

----if 1 gallon has 3.8 liters

then 12.6 gallons will have x liters

cross multiply we have

x= 12.6*3.8

x=47.88 liters

Hence the estimated amount in liters is 47.9 liters

8 0

3 years ago

3 + 10.8 is:<br> rational<br> irrational<br> Idk what this is if u could please help me

ratelena [41]

Answer:

3 + 10.8 is : rational.

Step-by-step explanation:

the reason for this is because it is neither a variable, pi, or numbers like square root 2.

4 0

2 years ago

The vertices of triangle RST are R (-7, 5), S(17, 5), T(5, 0). What is the perimeter of triangle RST?

MaRussiya [10]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ R(\stackrel{x_1}{-7}~,~\stackrel{y_1}{5})\qquad S(\stackrel{x_2}{17}~,~\stackrel{y_2}{5}) ~\hfill RS=\sqrt{(~~ 17- (-7)~~)^2 + (~~ 5- 5~~)^2} \\\\\\ ~\hfill RS=\sqrt{( 24)^2 + ( 0)^2}\implies \boxed{RS=24} \\\\\\ S(\stackrel{x_1}{17}~,~\stackrel{y_1}{5})\qquad T(\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) ~\hfill ST=\sqrt{(~~ 5- 17~~)^2 + (~~ 0- 5~~)^2}$

$~\hfill ST=\sqrt{( -12)^2 + ( -5)^2}\implies \boxed{ST=13} \\\\\\ T(\stackrel{x_1}{5}~,~\stackrel{y_1}{0})\qquad R(\stackrel{x_2}{-7}~,~\stackrel{y_2}{5}) ~\hfill TR=\sqrt{(~~ -7- 5~~)^2 + (~~ 5- 0~~)^2} \\\\\\ ~\hfill TR=\sqrt{( -12)^2 + (5)^2}\implies \boxed{TR=13} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{\LARGE perimeter}}{24~~ + ~~13~~ + ~~13\implies \text{\LARGE 50}}~\hfill$

3 0

1 year ago

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 <em>f</em> 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

3 years ago