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3 years ago

10

Find the range of the function y= 1/2 x + 3 when the domain is -2, 0, 2, 4

1 answer:

Mrrafil [7]3 years ago

7 0

Answer:

2 3 4 5

Step-by-step explanation:

1/2*-2+3 = 2

1/2*0+3 = 3

1/2*2+3 = 4

1/2*4+3 = 5

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Find the value of y in this equation 16y=164

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Read 2 more answers

Dada la ecuacion 25x2 + 4y2 = 100, determina las coordenadas de los vertices, focos, las longitudes de los respectivos ejes mayo

Likurg_2 [28]

Answer:

The given equation is

$25x^{2} +4y^{2}=100$

Which represents an elipse.

To find its elements, we need to divide the equation by 100

$\frac{25x^{2} +4y^{2} }{100} =\frac{100}{100} \\\frac{x^{2} }{4} +\frac{y^{2} }{25} =1$

Where $a^{2} =25$ and $b^{2}=4$ . Remember that the greatest denominator is $a$ , and the least is $b$ . So, we extract the square root on each equation.

$a=5$ and $b=2$ .

In a elipse, we have a major axis and a minor axis. In this case, the major axis is vertical and the minor axis is horizontal, that means this is a vertical elipse.

The length of the major axis is $2a=2(5)=10$ .

The length of the minor axis is $2b=2(2)=4$ .

The vertices are $(0,5);(0,-5)$ and $(2,0);(-2,0)$ .

Now, the main parameters of an elipse are related by

$a^{2}=b^{2} +c^{2}$ , which we are gonna use to find $c$ , the parameter of the focus.

$c=\sqrt{a^{2}-b^{2} }=\sqrt{25-4}=\sqrt{21}$

So, the coordinates of each focus are $(0,\sqrt{21})$ and $(0,-\sqrt{21})$

The eccentricity of a elipse is defined

$e=\frac{c}{a}=\frac{\sqrt{21} }{5} \approx 0.92$

The latus rectum is defined

$L=\frac{2b^{2} }{a}=\frac{2(4)}{5} =\frac{8}{5} \approx 1.6$

Finally, the graph of the elipse is attached.

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3 years ago