At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

May I ask what method of integration you chose?

first i used trig. sub. a-x^2, gives x=4sintheta, and then i used trig. integral of u-sub.

you could have done this with just an algebraic substitution

but it's x^3, not x^2

integral x^3*sqrt(16-x^2)dx

I missed an x in that first part but and that next equation you see i have x^2 times x which is x^3

expression* that follows that first equal sign

do you see?

Did you find P(x) as a 2nd degree taylor polynomial yet?

i do not know where to start......

\[P(x) \approx P(r)+P'(r)(x-r)+\frac{1}{2}P''(r)(x-r)^2 \]
This is what they want you to use.

so for 5, i got p5(x)=120000-40000(x-5)+22500(x-5)^2, is this right so far?

\[P(x)=45000+\frac{45000}{120000}(x-5)+\frac{1}{2} \frac{45000}{-40000}(x-5)^2 \]

That is how I got what was P'(5) and P''(5) and P(5)

We will use that P I wrote to approximate the value of portfolio bonds when r is 5.5.

oh are those commas?

lol

I can't read.

ok what you wrote is good then

So use your equation not mine to approximate P(5.5)

but how will it turn the equation into just 1 number?solve for x?

i don't understand

\[P(x) \approx 120000-40000(x-5)+45000(x-5)^2 \]
You can find P(5.5)
It will result in one number

there is only one variable

and you are asked to replace that variable x with 5.5

this will give you P(5.5)

It is just like if i asked you to evaluate f(2) given f(x)=x-5
you would say f(2)=2-5=-3

oh thank you so much, i thought i have to replace 5.5 in as in term of r

and don't forget the 1/2 part on that one part

but now i got it, 105625 would be the answer

yeah but you aren't given P''(5.5) or P'(5.5) or P(5.5) so that would be impossible

no that isn't the answer

don't forget the half part on that last term

\[P(x) \approx 120000-40000(x-5)+\frac{1}{2} 45000(x-5)^2 \]

which you wrote earlier has 22500 which is fine

I'm talking about for that 1/2*45000 part
you simplified it earlier to 22500