(Divide the following polynomials (x5 + y5) ÷ (x + y)

1 answer:

7 0

$\text{Use}\ a^5+b^5=(a + b) (a^4 - a^3 b + a^2 b^2 - a b^3 + b^4).\\\\(x^5+y^5):(x+y)=\dfrac{x^5+y^5}{x+y}=\dfrac{(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)}{x+y}\\\\=\boxed{x^4-x^3y+x^2y^2-xy^3+y^4}$

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2×10 to the six power times 0.00009

const2013 [10]

Answer:

5760

Step-by-step explanation:

(2 * 10)^6 * 0.00009

First solve inside the parentheses

20^6 * 0.00009

Then solve powers

64'000'000 * 0.00009

Solve multiplications

5760

This answer might not be the one that is asked for, as the writing is kind of vague, but if you turn this inn with explanation you will not get it wrong

5 0

2 years ago

Which is the standard form of the equation of the parabola that has a vertex of (3, 1) and a directric of x = -2?

vagabundo [1.1K]

Check the picture below. So,the parabola looks like so, notice the distance "p". since the parabola is opening to the right, then "p" is positive, thus is 5.

$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
\boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\
(x-{{ h}})^2=4{{ p}}(y-{{ k}}) \\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\$

$\bf -------------------------------\\\\
\begin{cases}
h=3\\
k=1\\
p=5
\end{cases}\implies (y-1)^2=4(5)(x-3)\implies (y-1)^2=20(x-3)
\\\\\\
\cfrac{1}{20}(y-1)^2=x-3\implies \boxed{\cfrac{1}{20}(y-1)^2+3=x}$