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Jada la siguiente función cuadrática \( y=2 x^{2}-6 x+4 \) a) Hallar: raices, vértice, eje de simetria y ordenada al origer b) Graficar

fx=x^2+2x

In Exercises 3-10, find the frequency of the function. \( \begin{array}{ll}\text { 3. } y=\sin x & \text { 4. } y=\sin 3 x \\ \text { 5. } y=\cos 4 x+2 & \text { 6. } y=-\cos 2 x \\ \text { 7. } y=\sin 3 \pi x & \text { 8. } y=\cos \frac{\pi x}{4} \\ \text { 9. } y=\frac{1}{2} \cos 0.75 x-8 & \text { 10. } y=3 \sin 0.2 x+6\end{array} \)

4. Find eigen values and eigen vectors of the matrix \( \left(\begin{array}{ll}5 & 1 \\ 4 & 2\end{array}\right) \) 5. Prove that \( r^{n} \vec{r} \) is an irrotational for any value of \( n \) but is solenoidal if \( n+3=0 \).

\( \begin{array}{l} \text { Dividiny Polynomials } \\ \left.x^{4}-9 x^{3}-3 x-14 x\right) \vdots(x-5)\end{array} \)

Ejercicios: \( \begin{array}{lll}\text { Derive las siguientes funciones con paréntesis: } \\ \begin{array}{lll}\text { a) } f(x)=\sqrt[3]{2 x+4} & \text { b) } f(x)=\sqrt[10]{x^{3}+10 x} & \text { c) } f(x)=\sqrt{x^{2}+3} \\ \begin{array}{lll}\text { d) } f(x)=\sqrt{1+\sqrt[3]{x}} & \text { e) } f(x)=\sqrt[4]{\sqrt{x}+\sqrt[3]{10 x}} & \text { f) } f(x)=\sqrt[3]{\sqrt{x}+3 x} \\ \text { j) } f(x)=\sqrt{x+\sqrt{x}+\sqrt[3]{x}} & \text { k) } f(x)=\sqrt[6]{x^{5}+\sqrt{x}} & \text { i) } f(x)=\sqrt{\sqrt{\sqrt{x+1}+7}}\end{array} \\ \end{array}\end{array} \$ \begin{array}{l}\text { 1) } f(x)=\sqrt[5]{\sqrt[3]{x^{2}+1}+7}\end{array} \)

1+1

(1 point) \( A \) is opposite side \( a, B \) is opposite side \( b \), and \( C \) is opposite side \( c \). Use the Law of Cosines to find angle \( A \) where \( a=45, b=17 \), and \( c=35 \). Round answer to the nearest tenth. \( A=1.3 \) help (angles)

Multiple Choice QuestionA problem in developing effective compensation for teams is that:rewarding all team members equally is most fairrewarding individuals builds cohesivenessrewarding individual performance is a new way to pay peoplerewarding individuals erodes cohesiveness

Example 5. For the differential equation \( y^{2}+x^{2} \frac{d y}{d x}=0 \), we found that the solution \( y(x) \equiv 0 \) was a singular point and this solution can not be obtain from the general solution \( y(x)=\frac{x}{(x-1)} \) by any choice of the constant \( c \).