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Lady_Fox [76]
3 years ago
13

Use Euler's method to obtain a four-decimal approximation of the indicated value. First use h = 0.1 and then use h = 0.05.

Mathematics
1 answer:
Juliette [100K]3 years ago
4 0

Answer:

  see below for the tables

Step-by-step explanation:

The differential equation is separable, so the solution is ...

  \displaystyle\dfrac{dy}{dx}=2xy\\\\\int{\dfrac{dy}{y}}=\int{2x}\,dx\\\\\ln{y}=x^2+C\\\\\text{Considering the initial condition, $C=-1$}\\\\\boxed{y=e^{x^2-1}}

__

The values for yn are y+y'·h = y+2xyh. We take the "absolute error" to be the (signed) difference between the calculated yn and the actual value y(x).

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Stephon has a square brick patio. He wants to reduce the width by 4 feet and increase the length by 4 feet.
alekssr [168]

Answer:  The expressions for the length and width of the new patio are

\ell=x+4,~~w=x-4.

And the area of the new patio is 384 sq. feet.

Step-by-step explanation:  Given that Stephen has a square brick patio. He wants to reduce the width by 4 feet and increase the length by 4 feet.

The length of one side of the square patio is represented by x.

We are to write the expressions for the length and width of the new patio and then to find the area of the new patio if the original patio measures 20 feet by 20 feet.

Since Stephen wants to reduce width of the patio by 4 feet, so the width of the new patio will be

w=(x-4)~\textup{feet}.

The length of the patio is increased by 4 feet, so the length of the new patio will be

\ell=(x+4)~\textup{feet}.

Now, if the original patio measures 20 feet by 20 feet, then we must have

w=x-4=20-4=16~\textup{feet}

and

\ell=x+4=20+4=24~\textup{feet}.

Therefore, the area of the new patio is given by

A_n=\ell \times w=24\times16=384~\textup{sq. feet}.

Thus, the expressions for the length and width of the new patio are

\ell=x+4,~~w=x-4.

And the area of the new patio is 384 sq. feet.

4 0
4 years ago
Simplify each expression:<br> a. 32=(-7+5)<br> b. 18+42 =(-8)
Sedbober [7]
Unfortunately these aren’t expressions, these are equations that are not equivalent.
6 0
3 years ago
In a recent 10 year period , the change in the number of visitors to an amusement park was about 15,000 visitors. What was the m
Andrej [43]
1,500 was the mean yearly change
7 0
4 years ago
Ms. Pacheco, Mr Edwards, and Mr Richards are three math teachers at turner middle school. Ms. Pacheco is three years older than
zubka84 [21]
Mr edwards = Mr r x 2
Mr r = Mr p - 3 = Mr e ÷ 2.                       Mr p = 30
Mr e + Mr r = 81.                                       Mr e = 54
Mr p = Mr r + 3.                                         Mr r = 27

Mr Edwards = 2(Mr p - 3)
Mr e = 2(mr p) - 6

Mr r = [2(Mr p) -6] ÷ 2
Mr r = [2( Mr r + 3) - 6] ÷ 2

2(Mr r) + (Mr p - 3) = 81
2(Mr p -3) + (Mr p - 3) = 81
3(Mr p - 3) = 81 
3(Mr p) - 9 = 81.                               Hope this helps!!!
3(Mr p) + 9 = 81 + 9
3(Mr p) = 90
Mr p = 90 ÷ 3
Mr p = 30


5 0
3 years ago
Read 2 more answers
5x-4y-6z = -3<br> X-3y+z=-1<br> -3x-6y+7z=1
PilotLPTM [1.2K]

Answer:

x=-\frac{41}{55},y=\frac{2}{55},z=-\frac{8}{55}

Step-by-step explanation:

5x-4y-6z=-3...............eq(1)\\\\x-3y+z=-1.....................eq(2)\\\\-3x-6y+7z=1.................eq(3)

eq(1)-5\times eq(2)

5x-4y-6z-5(x-3y+z)=-3-5\times (-1)\\\\11y-11z=2\\\\y-z=\frac{2}{11}...................eq(4)

3\times eq(2)+eq(3)

3(x-3y+z)-3x-6y+7z=3\times (-1)+1\\\\-15y+10z=-2\\\\-y+\frac{2}{3}z=-\frac{2}{15}.........eq(5)

eq(4)+eq(5)

y-z-y+\frac{2}{3}z=\frac{2}{11}-\frac{2}{15}\\\\-\frac{1}{3}z=\frac{8}{165}\\\\z=-\frac{8\times 3}{165}\\\\z=-\frac{8}{55}

From eq(4)

y=z+\frac{2}{11}=-\frac{8}{55}+\frac{2}{11}\\\\y=\frac{2}{55}

Substitute the value of y and z in eq(2)

x-3\times \frac{2}{55}-\frac{8}{55}=-1\\\\x-\frac{14}{55}=-1\\\\x=-1+\frac{14}{55}\\\\x=\frac{41}{55}

6 0
4 years ago
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