Choose the polynomial that is written in standard form? |||||Pictured|||||

Mathematics

2 answers:

Usimov [2.4K]3 years ago

8 0

The 3rd one down is correct

-Dominant- [34]3 years ago

6 0

Answer:

C. $-3x^8y^2+9x^2y+10x^2$

Step-by-step explanation:

We are asked to choose the polynomial that is written in standard form.

We know that a polynomial is written in standard form, when it is written in descending order of degrees that is from highest degree to lowest degree.

Upon looking at our given choices, we can see that option C is the correct choice as it is written in descending order of degrees.

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Find dy/dx x=a(cost +sint) , y=a(sint-cost)

MissTica

Answer:

$\begin{aligned} \frac{dy}{dx} &= \frac{\cos(t) + \sin(t)}{\cos(t) - \sin(t)} \end{aligned}$ given that $a \ne 0$ and that $\cos(t) - \sin(t) \ne 0$ .

Step-by-step explanation:

The relation between the $y$ and the $x$ in this question is given by parametric equations (with $t$ as the parameter.)

Make use of the fact that:

$\begin{aligned} \frac{dy}{dx} = \quad \text{$\frac{dy/dt}{dx/dt}$ given that $\frac{dx}{dt} \ne 0$} \end{aligned}$ .

Find $\begin{aligned} \frac{dx}{dt} \end{aligned}$ and $\begin{aligned} \frac{dy}{dt} \end{aligned}$ as follows:

$\begin{aligned} \frac{dx}{dt} &= \frac{d}{dt} [a\, (\cos(t) + \sin(t))] \\ &= a\, (-\sin(t) + \cos(t)) \\ &= a\, (\cos(t) - \sin(t))\end{aligned}$ .

$\begin{aligned} \frac{dx}{dt} \ne 0 \end{aligned}$ as long as $a \ne 0$ and $\cos(t) - \sin(t) \ne 0$ .

$\begin{aligned} \frac{dy}{dt} &= \frac{d}{dt} [a\, (\sin(t) - \cos(t))] \\ &= a\, (\cos(t) - (-\sin(t))) \\ &= a\, (\cos(t) + \sin(t))\end{aligned}$ .

Calculate $\begin{aligned} \frac{dy}{dx} \end{aligned}$ using the fact that $\begin{aligned} \frac{dy}{dx} = \text{$\frac{dy/dt}{dx/dt}$ given that $\frac{dx}{dt} \ne 0$} \end{aligned}$ . Assume that $a \ne 0$ and $\cos(t) - \sin(t) \ne 0$ :

$\begin{aligned} \frac{dy}{dx} &= \frac{dy/dt}{dx/dt} \\ &= \frac{a\, (\cos(t) + \sin(t))}{a\, (\cos(t) - \sin(t))} \\ &= \frac{\cos(t) + \sin(t)}{\cos(t) - \sin(t)}\end{aligned}$ .