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DaniilM [7]
3 years ago
7

Brainliest answer to the right answer and 30 points!

Mathematics
1 answer:
timofeeve [1]3 years ago
8 0
10.5^2 + 20.8^2 = 110.25 + 432.64 = 542.89
23.3^2 = 542.89 
Right triangle
------------------------
20.1^2 = 524.41
6^2 + 20.1^2 = 36 + 404.01 = 440.01
Not a right triangle
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What’s the difference between a radius and a diameter?
otez555 [7]

Answer:

radius is half of the circle. diameter is the line that goes all the way to the other end of the circle

Step-by-step explanation:


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3 years ago
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Which answer shows 309,070 written in word form? A. three hundred nine thousand, seven hundred B. thirty-nine thousand, seventy
aleksley [76]

The correct answer is D

Hope this helps

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3 years ago
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The cost of parking a motor home in a state park includes a one time fee of $180, plus a daily rate of $55 per day. Anthony prep
HACTEHA [7]

Answer:

Anthony paid for 7 days of parking.

Step-by-step explanation:

From the information given, the cost of parking a motor home in a state park would be equal to the one time fee plus the result of multiplying the daily rate for the number of days:

c=180+55x, where c is the cost of parking the motor home and x is the number of days.

Now, you can replace c with 565 that is the amount Anthony prepaid and you can solve for x to be able to find the number of days he paid for:

565=180+55x

565-180=55x

385=55x

x=385/55

x=7

According to this, Anthony paid for 7 days of parking.

7 0
2 years ago
Find the area of figure 1.<br><br> a -12 units2<br> b- 14 units2<br> c- 8 units2<br> d- 10 units2
Brilliant_brown [7]

Answer:

D

Step-by-step explanation:

if you look at the chart. it goes up five and has a width of 2. multiply 5 x 2 = you get 10

3 0
2 years ago
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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 <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
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