Linear functions m and p are shown

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

The data set below shows the number of players on each softball team in a tournament:

stich3 [128]

An outlier number is the number that's completely different from the rest.
For example;
1, 15, 17, 16, 14.
"1" is the outlier.

Our list of numbers range (on average) from 7-12.

Let's look at our answer choices.

A.) There is one outlier that indicates an unusually large number of players on that team.

This is true, as we have 21, the one and only outlier in our list.

Your answer is A.)

I hope this helps!

7 0

3 years ago

What point is on the graph of every direct variation equation

Alexeev081 [22]

Answer: The point (0,0)
This point is also known as the origin

The reason why is because every direct variation equation is of the form y = k*x
where k is some fixed number

If we plugged in x = 0 it leads to y = k*x = k*0 = 0. So (x,y) = (0,0)

4 0

3 years ago

Can someone help me. I can’t started on how to figure this out <br> Thank you

salantis [7]

Answer: for the first one is the second one

for the next one is the last one

3 0

2 years ago

Find the domain of the relation<br> find the range of the relation<br> is the relation a function?

iren2701 [21]

Answer:

Domain: (-∞ , ∞)

Range: [4 , -∞)

yes, it's a function. It pass the vertical line test. (one input vs one output)

Step-by-step explanation:

Domain: (-∞ , ∞)

Range: [4 , -∞)

yes, it's a function. It pass the vertical line test. (one input vs one output)

6 0

3 years ago