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inessss [21]
3 years ago
10

Please do the agree or disagree part and justification

Mathematics
1 answer:
Jlenok [28]3 years ago
8 0

Answer:

<em>We disagree with Zach and Delia and agree with Alicia</em>

Step-by-step explanation:

The domain of a function is the set of values of the independent variable that the function can take according to given rules or restrictions.

The range is the set of values the dependent variable can take for every possible value of the domain.

The graph shows a continuous line representing the values of the function. We must take a careful look to the values of x (horizontal axis) where the function exists. It can be done by drawing an imaginary vertical line passing through the value of x. If that line touches the graph of the function, it belongs to the domain. It's clear that every value of x between -5 and 3 (both inclusive because there are solid dots in the extremes) belong to the domain:

Domain: -5\leq x \leq 3

The range is obtained in a similar way as the domain, but the imaginary lines must be horizontal. That gives us the values of y range from -7 to 5 both inclusive:

Range:

-7\leq y \leq 5

Thus we disagree with Zach and Delia and agree with Alicia

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Using the method of completing the square, put each circle into the form
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Answer:

Standard form: (x-\frac{1}{2})^2 + (y-0)^2 = 15

Center: (\frac{1}{2}, 0)

Radius: r =\sqrt{15}

Step-by-step explanation:

The equation of a circle in the standard form is

(x-h)^{2} + (y-k)^{2} = r^{2}

Where the point (h, k) is the center of the circle

To transform this equation 4x^{2} -4x + 4y^{2} - 59 = 0 this equation  in the standard form we use the method of square.

First, we group similar variables

(4x^{2} -4x) + (4y^{2}) - 59 = 0

Divide both sides of equality by 4

(x^{2} -x) + (y^{2}) - 14.75 = 0

Now we complete square for variable x.

Take the coefficient "b" that accompanies the variable x and divide by 2. Then, elevate the result to the square:

b =-1\\\\\frac{b}{2}= \frac{-1}{2}= -\frac{1}{2}\\\\(\frac{b}{2})^2=  (-\frac{1}{2})^2 = \frac{1}{4}

Now add (\frac{b}{2})^2 on both sides of the equality

(x^{2} -x +\frac{1}{4}) + (y^{2}) - 14.75 = (\frac{1}{4})

Factor the expression and simplify the independent terms

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(x-\frac{1}{2})^2 + (y-0)^2 = 15

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h =\frac{1}{2}\\\\k=0

and the center is (\frac{1}{2}, 0)

radius r =\sqrt{15}

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