What 2 things are recommended in order for you to have a neat graph

Mathematics

1 answer:

Oksana_A [137]3 years ago

6 0

Answer:

use a ruler

buy a graph

Step-by-step explanation:

Not the best answer tho... becuase this useless Mod deleted all my answers randomly

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If f(x) = kx^3+ x^2 − kx + 2, find a number k such that the graph of f contains the point (k, -10)

Elina [12.6K]

If we write $k$ where we see $x$ in the equation and set the result equal to $-10$ , we get the result.

$f(k)=k(k)^3+k^2-k(k)+2$
$=k^4+k^2-k^2+2$
$=k^4+2=-10$
$k^4=-12$
$k_{1}=\sqrt[4]{3}(-1-i)$
$k_{2}=\sqrt[4]{3}(-1+i)$
$k_{3}=\sqrt[4]{3}(1-i)$
$k_{4}=\sqrt[4]{3}(1+i)$

4 0

1 year ago

Question:

Tamiku [17]

Answer:

Step-by-step explanation:

$3x^2 + 12x + 6 = 0\\\\a=3\,,\ b=12\,,\ c=6\\\\\\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ x=\dfrac{-12\pm\sqrt{12^2-4\cdot3\cdot6}}{2\cdot3}=\dfrac{-12\pm\sqrt{144-72}}{2\cdot3}=\dfrac{-12\pm\sqrt{72}}{2\cdot3}=\dfrac{-12\pm6\sqrt{2}}{6}\\\\x_1=\dfrac{6(-2+\sqrt{2})}{6}=-2+\sqrt2\\\\x_2=\dfrac{6(-2-\sqrt{2})}{6}=-2-\sqrt2$