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andreyandreev [35.5K]
2 years ago
15

Easy make sure its detailed 15 points

Mathematics
1 answer:
goblinko [34]2 years ago
4 0

9514 1404 393

Answer:

  • no square roots: -1000, -8
  • one square root: 0
  • two square roots: 8, 64, 1000
  • no cube roots: <none>
  • one cube root: -1000, -8, 0, 8, 64, 1000
  • two cube roots: <none>

Step-by-step explanation:

The attached graph shows the square root relation (red) and the cube root function (blue). The function values are shown for x=0 and x=±8.

You can see that there are 2 square roots for positive numbers, one square root for 0, and 0 square roots for negative numbers. There is exactly 1 cube root for any number.

  • no square roots: -1000, -8
  • one square root: 0
  • two square roots: 8, 64, 1000
  • no cube roots: <none>
  • one cube root: -1000, -8, 0, 8, 64, 1000
  • two cube roots: <none>

_____

<em>Additional comment</em>

We call the square root curve a "relation" because it is <em>not a function</em>. A relation that is a function will have only one y-value for each x-value. For positive x-values, there are two square roots.

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Outline the process for Inscribing an equilateral triangle in a circle. Perform the construction in GeoGebra, and take a
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Answer:

I don't use Geogebra, but the following procedure should work.

Step-by-step explanation:

Construct a circle A with point B on the circumference.

  1. Use the POINT and SEGMENT TOOLS to create a circle with centre B and radius BA.
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Answer:

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Before Simplifying:

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a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l
GrogVix [38]

Answer:

\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}

Step-by-step explanation:

The equation of a line:

y=mx+b

We have

m=\dfrac{2}{3}

substitute:

y=\dfrac{2}{3}x+b

The formula of a distance between a point and a line:

General form of a line:

Ax+By+C=0

Point:

(x_0,\ y_0)

Distance:

d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}

Convert the equation:

y=\dfrac{2}{3}x+b     |<em>subtract y from both sides</em>

\dfrac{2}{3}x-y+b=0    |<em>multiply both sides by 3</em>

2x-3y+3b=0\to A=2,\ B=-3,\ C=3b

Coordinates of the point:

(0,\ 0)\to x_0=0,\ y_0=0

substitute:

d=4

4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}

4=\dfrac{|3b|}{\sqrt{13}}\qquad|    |<em>multiply both sides by \sqrt{13}</em>

4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}   |<em>divide both  sides by 3</em>

b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}

Finally:

y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}

4 0
2 years ago
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