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2 years ago

Find the area of the complex figure

Mathematics

1 answer:

Charra [1.4K]2 years ago

4 0

Area should be 60 cm squared.

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What is 80%of 1,60<br>will mark brainliest

cluponka [151]

Answer:

Step-by-step explanation:

Converte 80% into a decimal (devide it by 100) = 0.8

Then multiply 0.8 by 1.60 = 1.28 (final result)

4 0

2 years ago

If you can get an answer to any question, what would you ask? You toss a fair coin 4 times. What is the probability that (round

Andrew [12]

Answers:

a) 0.0625

b) 0.9375

==================================================

Work Shown:

The probability of landing on heads is 1/2 = 0.5 since both sides are equally likely to land on. Getting 4 heads in a row is (1/2)^4 = (0.5)^4 = 0.0625

The event of getting at least one tail is the complement of getting all four heads. This is because you either get all four heads or you get at least one tail. One or the other must happen. We subtract the result we got from 1 to get 1-0.0625 = 0.9375

You can think of it like this

P(getting all four heads) + P(getting at least one tail) = 1

The phrasing "at least one tail" means "one tail or more".

4 0

2 years 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

What's the slope of (-2,-1) and (4,3)

Llana [10]

Answer:

4/6

Step-by-step explanation:

→ Write the slope formula

(y₂ - y₁) ÷ (x₂ - x₁)

→ Substitute in the numbers

(3 - -1) ÷ (4 - -2)

→ Simplify

4/6

5 0

3 years ago

Read 2 more answers

Math: hey everyone, can anyone help me find the answer for this math question? I don’t quite understand but yea the picture is t

Karo-lina-s [1.5K]

Answer:

4 1/3

I hope this helps

6 0

2 years ago