Helpppp asappp! Pleaseeeeeeeeeeee four questions!! Thank you!!

[susu]

1. Find a mathematical model representing the statement. (Determine the constant of proportionality.)
P varies directly as x and inversely as the square of y. P=28/3, when x=56 and y=12)
. Okay for this one i just tried using the inverse variation. and manipulating the formula. I got 2x over y. But it's wrong. Someone help me please.


2.Write an equation for the parabola in standard form. The points on the graph are (-1,-2) (0,-1)
. I tried to use the ax^2+bx+c formula than just switching it to the a(x-h)^2+k formula but my answer is getting rejected.


3. Find the values of b such that the function has the given minimum value.

f(x)= x^2+bx-20; Minimum Value: -29

I have solved this one more than 7times! I got 10, -10. But i have no clue what i'm doing wrong. IF someone could work this one out entirely that would be lovely.

 

[galactus]

Your first one is close, you just forgot the square on your y.

\(\displaystyle P=\frac{kx}{y^{2}}\), not y.

For #2, you usually need three points to find a parabola equation. Is there more info you left out?.

From (-1,2):

\(\displaystyle a(-1)^{2}-b+c=-2\)

and from (0,-1): \(\displaystyle c=-1\)



For the third problem, the derivative is \(\displaystyle f'(x)=2x+b\)

So, we also have \(\displaystyle x^{2}+bx-20=-29\)

Two equations with two unknowns. It is not 10.

 

[susu]

Thank you but it still tells me that the first one is wrong

 

[galactus]

You have to find k given the values.

\(\displaystyle 28/3=\frac{k(56)}{144}\)

solve for k.

Then, it goes into your formula.

 

[susu]

oh yes thank you! lol. There's one more question i've been doing for a few minutes kind of lost .
Two polynomials P and D are given. Use either synthetic or long division to divide P(x) by D(x), and express the quotient P(x)/D(x) in the form
P(x)/D(x)=Q(x)+R(x)/D(x).. P(x)= x^5+x^4-5x^3+4x+5, D(x)=x^2+x-1


P(x)/D(x)=?


I know both synthetic division and long division but i am SO confused with this one. I regularly divided it but i don't know what to do next

1

 

[Deleted member 4993]

1. Find a mathematical model representing the statement. (Determine the constant of proportionality.)
P varies directly as x and inversely as the square of y. P=28/3, when x=56 and y=12)
. Okay for this one i just tried using the inverse variation. and manipulating the formula. I got 2x over y. But it's wrong. Someone help me please.


2.Write an equation for the parabola in standard form. The points on the graph are (-1,-2) (0,-1)
. I tried to use the ax^2+bx+c formula than just switching it to the a(x-h)^2+k formula but my answer is getting rejected.


3. Find the values of b such that the function has the given minimum value.

f(x)= x^2+bx-20; Minimum Value: -29

I have solved this one more than 7times! I got 10, -10. But i have no clue what i'm doing wrong. IF someone could work this one out entirely that would be lovely.

oh yes thank you! lol. There's one more question i've been doing for a few minutes kind of lost .
Two polynomials P and D are given. Use either synthetic or long division to divide P(x) by D(x), and express the quotient P(x)/D(x) in the form
P(x)/D(x)=Q(x)+R(x)/D(x).. P(x)= x^5+x^4-5x^3+4x+5, D(x)=x^2+x-1


P(x)/D(x)=?


I know both synthetic division and long division but i am SO confused with this one. I regularly divided it but i don't know what to do next

You have been posting problems - looks like for a take home test - without showing a line of work.

Please provide the e-mail address of your teacher so that we can confitm that it is permissible to ask for help (without your work) for these problems.

 

[soroban]

Hello, susu!

Since this is posted under "Intermediate Algebra",
. . I will assume you are not familiar with Calculus.

3. Find the values of . \(\displaystyle b\) such that the function has the given minimum value.

. . \(\displaystyle f(x)\:=\: x^2+bx-20,\;\text{ minimum value: }-29\)

The graph of the equation is an up-opening parabola.
The minimum point is at its vertex.

The formula for the vertex of the parabola \(\displaystyle y \:=\:ax^2 + bx + c\)
. . is given by: .\(\displaystyle x \,=\,\dfrac{\text{-}b}{2a}\)

Our parabola has: .\(\displaystyle a = 1,\;b = b,\;c = \text{-}20\)

. . Hence, the vertex is at: .\(\displaystyle x \,=\,\text{-}\dfrac{b}{2}\)


The \(\displaystyle y\)-coordinate of the vertex is: .\(\displaystyle y \:=\:\left(\text{-}\frac{b}{2}\right)^2 + b\left(\text{-}\frac{b}{2}\right) - 20\)
. . \(\displaystyle y \:=\:\text{-}\frac{b^2}{4} - 20\:\text{ which equals -29.}\)

Hence, we have: .\(\displaystyle \text{-}\dfrac{b^2}{4} - 20 \:=\:\text{-}29 \quad\Rightarrow\quad \dfrac{b^2}{4} \:=\:9 \quad\Rightarrow\quad b^2 \:=\:36 \)

. . Therefore: .\(\displaystyle b \:=\:\pm\,6\)