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1 year ago

Need help with this math.

Mathematics

1 answer:

madreJ [45]1 year ago

3 0

The end behavior of the function y = x² is given as follows:

f(x) -> ∞ as x -> - ∞; f(x) -> ∞ as x -> - ∞.

<h3>How to identify the end behavior of a function?</h3>

The end behavior of a function is given by the limit of f(x) when x goes to both negative and positive infinity.

In this problem, the function is:

y = x².

When x goes to negative infinity, the limit is:

lim x -> - ∞ f(x) = (-∞)² = ∞.

Meaning that the function is increasing at the left corner of it's graph.

When x goes to positive infinity, the limit is:

lim x -> ∞ f(x) = (∞)² = ∞.

Meaning that the function is also increasing at the right corner of it's graph.

Thus the last option is the correct option regarding the end behavior of the function.

<h3>Missing information</h3>

We suppose that the function is y = x².

More can be learned about the end behavior of a function at brainly.com/question/24248193

#SPJ1

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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

GuDViN [60]

Answer:

Euler's method is a numerical method used in calculus to approximate a particular solution of a differential equation. As a numerical method, we have to apply the same procedure many times, until get the desired result.

In first place, we need to know all the values the problem is giving:

The step size is 0.2; h = 0.2. This step size is a periodical increase of the x-variable, which will allow us to calculate each y-value to each x.
The problem is asking the solution y(1), which means that we have to find the y-value assigned for x = 1, through the numerical method.
The initial condition is y(0) = 9. In other words, $x_{o} = 0\\y_{0}=9$ .

So, if the initial x-value is 0, and the step size is 0.2, the following x-value would be: $x_{1}=0.2$ ; then $x_{2}=0.4$ ; $x_{3} =0.6; x_{4} =0.8;x_{5} =1$ ; and so on.

Now, we have to apply the formula to find each y-value until get the match of $x_{5}=1$ , because the problem asks the solution y(1).

According to the Euler's method:

$y_{1} =y_{0} +hF(x_{0};y_{0})\\y_{2} =y_{1} +hF(x_{1};y_{1})\\y_{n} =y_{n-1} +hF(x_{n-1};y_{n-1})$

Where $F(x;y)=x^{2} y-12y^{2}$ , and $x_{0} =0; y_{0} =9$ ; $h=0.2$ .

Replacing all values we calculate the y-value assigned to $x_{1}$ :

$y_{1} =9+0.2((0)^{2} 9-12(9)^{2})=-185.4$ .

Now, $y_{1} =-185.4$ , $x_{1} =0.2; h=0.2$ . We repeat the process with the new values:

$y_{2} =y_{1} +hF(x_{1};y_{1}) \\y_{2} = -185.4+0.2((0.2)^{2} (-185.4)-12(-185.4)^{2} )\\y_{2}=-82682.47$

Then, we repeat the same process until get the y-value for $x_{5} =1$ , which is $y_{5} = -1.0018$ , round to four decimal places.

Therefore, $y(1)=-1.0018$ .

7 0

3 years ago

Will mark Brianliest please answer :)

Brilliant_brown [7]

Answer:

3.5×10^4=35000

Step-by-step explanation:

Consider the value

3.5

we are saying that we are counting in units (1's) in that we have

of 1's

'.....................................................................................

Consider the value

3.5

. Now we are saying we have

of tens. That is; 3 lots of 10 is 30 and

of 10 is 5 so we have

'.................................................................................

Note that

is 100

Consider the value

3.5

. Now we are saying we have

100

So if we are counting in 100's then the decimal point stays where

it is and we move the number of 3.5 to the left until the

3 becomes hundreds.

Note that we have to insert a place holder of 0 just to the left of the decimal point to make sure the three is viewed as in the hundreds.

3.5

−−−−−−−−−−−−−−−−−−→

slide to the left 2 places

350.0

6 0

3 years ago

How far above ground is the top of the pole?

Arisa [49]

Answer:

I would say about 7 feet and 1 inch

Step-by-step explanation:

8 0

3 years ago

Help please asappppppppp

Aleonysh [2.5K]

Answer:

First Equation: 6a-23

Second Equation: 5a-20

Step-by-step explanation:

So, I think you want me to answer part 2...

For the first equation, we can see that the denominator stays the exact same, meaning the numerator will also not change.

For the second equation, we can see the denominator changed from a-3 to 5a-15. The denominator was simply multiplied by 5. Since the denominator was multiplied by five we must multiply the numerator by five as well.

5(a-4)

5a-20

Hope this helps :)

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IrinaVladis [17]

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