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xxTIMURxx [149]
3 years ago
9

Using the graph, determine the coordinates of the roots of the parabola​

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
6 0

Answer:

(-1, 4)

Step-by-step explanation:

Use of the coordinate plane.

The vertex of the parabola is 1 to the left and up 4.

Think of the vertex as where the curve starts.

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1. Which of the following is true regarding the graph of y = 1/x ?
KATRIN_1 [288]

Answer:

b. There are horizontal and vertical asymptotes

Step-by-step explanation:

The graph is a hyperbola with asymptotes of x=0 and y=0

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Is 5.78778778 a rational or irrational number? Explain
Bumek [7]
Hello There!

It is rational as it can be written as a fraction and had a pattern to it.

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18 Ib. Then her dog loses another 1 lb. Finally, it
Pavel [41]

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17 lb I guess 8 don't think the question is full

Step-by-step explanation:

the question is half

8 0
3 years ago
Which expression is equivalent to (4^−3)^−6?<br><br>1. 4^-18<br>2. 4^18<br>3. 4^-9<br>4. 4^3
malfutka [58]

Answer:

4^18

Step-by-step explanation:

4^18 = 2^36

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Please visit: https://youtu.be/_0Ft9HdNg5c

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8 0
3 years ago
The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the
Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

7 0
3 years ago
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