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Afina-wow [57]
2 years ago
15

Which notation indicates 2 congruent segments

Mathematics
1 answer:
yaroslaw [1]2 years ago
6 0

Answer:

≅ is congruent

Step-by-step explanation:

║ is parallel

⊥ perpendicualr

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Consider the ODE, dy dx = y 2 1 + x (2) subject to condition y = 1 when x = 0, use your Euler code from class (modified if neces
makkiz [27]

Answer:

Computation.

Step-by-step explanation:

I'm not really sure if that's the analytical solution of the inital value problem,

because y(0)=11-ln(1-0)(3)=11. Howevwer, let us procede with the given values...

Let us assume that we are going to use euler with n=2 (two steps) and h=0.2(the size of each step)

The update rules of the Euler Methode are

X_i = X_{i-1}+h=X_0+ih

m_i=\dfrac{dY}{dX} \biggr \rvert_{x_i} \\\\Y_{i+1}=m_i\cdot h+y_i

Since the initial value problem tells us that Y=1 when X=0, we know that

X_0=0 and that Y_0=1. Then, we have

X_0=0\\\\X_1=0.2\\\\X_2=0.4

and

Y_0=1\\\\m_0=21 \cdot Y_0 + X_0 \cdot 2=21 \cdot 1 + 0=21\\\\Y_1=0.2 \cdot 21 + 1 =5.2\\\\m_1=21 \cdot Y_1 + X_1\cdot 2=21 \cdot 5.2 + 0.2 \cdot 2=109.6\\ \\Y_2=m_1 \cdot h + Y_1 = 109.6 \cdot 0.2 + 5.2= 27.12

which gives us the points (0,1), (0.2, 5.2) and (0.4, 27.12).

Now, since we want to compare the analyticaland the Euler result, we first compute the value of y=11-ln(1-x)(3) for the values x=0, 0.2 and 0.4. We get that

y(0)=11-\ln(1-0)(3)=11-ln(1)(3)=11\\\\Y(1)=11-\ln(1-0.2)(3)=11.67\\\\Y(2)=11-\ln(1-0.4)(3)=12.53

and we compute Y(i)-Y_i for each i.

It holds

Y(0)-Y_0=11-1=10\\\\Y(1)-Y_1=11.67-5.2=6.47\\\\Y(2)-Y_2=12.53-27.12=-14.59

which tells us that we have a really bad approximation, as I already stated there must be a mistake in the analytical solution since the intial values don't coincide. Also note that the curve that we get using the euler methose is growing faster than the analitical solution.

4 0
3 years ago
Find the 6th term of a geometric sequence t3 = 444 and t7 = 7104. So I have to find r. But is this right: 7104 = 444r^4 r^4 = 16
galben [10]

Third term = t3 = ar^2 = 444           eq. (1)

Seventh term = t7 = ar^6 = 7104         eq. (2)

By solving (1) and (2) we get,

              ar^2 = 444    

                => a = 444 / r^2       eq. (3)

And  ar^6 = 7104

 (444/r^2)r^6 = 7104

 444 r^4 = 7104

 r^4 = 7104/444

            = 16

 r2 = 4

 r = 2

Substitute r value in (3)

                         a = 444 / r^2

                             = 444  / 2^2

                             = 444 / 4

                              = 111

Therefore a = 111 and r = 2

Therefore t6 = ar^5

                       = 111(2)^5

                       = 111(32)

                       = 3552.

<span>Therefore the 6th term in the geometric series is 3552.</span>

8 0
3 years ago
If 4 dice are rolled, what is the number of ways in which at least 1 die shows 3?
UkoKoshka [18]

I think the number of ways is 671?

4 0
3 years ago
HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPp
love history [14]

Answer:

<u>36 in²</u>

Step-by-step explanation:

<u>Area of the figure</u>

  • Area (small rectangle) left + Area (big rectangle) right
  • 3 x (6 - 4) + 5 x 6
  • 3 x 2 + 30
  • 6 + 30
  • <u>36 in²</u>

3 0
2 years ago
Read 2 more answers
Help me pls pls pls ​
Simora [160]

Answer:

A.the product of X and a factor not depending on X.

4 0
3 years ago
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