Calculate. $y=3{x}^{5}+5{x}^{4}-2{x}^{3}$

Noelanijd
2022-07-27
Answered

Calculate. $y=3{x}^{5}+5{x}^{4}-2{x}^{3}$

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A lattice point in the xy-plane is a point both of whose coordinates are integers (not necessarily positive). How many lattice points lie on the hyperbola ${x}^{2}-{y}^{2}=17$ ?

I think the answer should be 4, because${x}^{2}-{y}^{2}=(x+y)(x-y)=17$ . 17 has 4 factors: 1,17, -1, and -17. But I don't know if these numbers actually work.

I think the answer should be 4, because

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Question

Solve the given quadratic eqauation by using the graph of its related function.

${x}^{2}-67=-51$

Part 1 out of 2

Enter te related function of the given quadratic equation. Then complete the table of values for the related function.

The related function is$y={x}^{2}+$ .

$$\begin{array}{|cccccc|}\hline x& -5& -4& 0& 4& 5\\ y\\ \hline\end{array}$$

Solve the given quadratic eqauation by using the graph of its related function.

Part 1 out of 2

Enter te related function of the given quadratic equation. Then complete the table of values for the related function.

The related function is

asked 2022-08-05

1.Show that $\overline{{z}_{1}+{z}_{2}}=\overline{{z}_{1}}+\overline{{z}_{2}}$

2.Use Exercise 1 to show that $\overline{{z}^{2}}=(\overline{z}{)}^{2}$

the z=a+bi

$\overline{z}=a-bi$

$\overline{{z}_{1}}=a+bi$

2.Use Exercise 1 to show that $\overline{{z}^{2}}=(\overline{z}{)}^{2}$

the z=a+bi

$\overline{z}=a-bi$

$\overline{{z}_{1}}=a+bi$

asked 2022-03-23

Making sense of William Jones's solution of quadratic equations and notation

William Jones, when discussing quadratic equations, says: "Therefore if$\vee$ be put for the Sign of any Term, and ∧ for the contrary, all Forms of Quadratics with their Solutions, will be reduc'd to this one. If $xx\vee ax\vee b=0$ then $\wedge \frac{1}{2}a\stackrel{\u2015}{\pm aa\wedge b}!,{\mid}^{\frac{1}{2}}$ ."

Could you please help me make sense:

(a) What is meant with "the contrary" of "any sign"?

(b) Do I understand correctly that the "then" part supposed to be a solution to the preceding quadratic equation? If yes, how does this work exactly? How does it align with how we would symbolize the solution today?

William Jones, when discussing quadratic equations, says: "Therefore if

Could you please help me make sense:

(a) What is meant with "the contrary" of "any sign"?

(b) Do I understand correctly that the "then" part supposed to be a solution to the preceding quadratic equation? If yes, how does this work exactly? How does it align with how we would symbolize the solution today?