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maria [59]
3 years ago
7

What is the domain: What is the range:

Mathematics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:

Domain: [-4,4]

Range: [-4,6]

Step-by-step explanation:

To domain pertains to the x-coordinates, while the range pertains to the y-coordinates.

Looking at the graph, we can see that the farthest point to the left is (-4,3) and the farthest point to the right is (4,6). Now, we know that the domain is [-4,4].

Looking at the graph, we can see the highest point is (4,6) and the lowest point is (0,-4). Now, we know the range is [-4,6].

Now, we know that the domain is [-4,4] and the range is [-4,6].

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Perimeter of a triangle with side lengths 2x, (3x-1) and (4x+5)?
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Does is have a total perimeter of the triangle? If not, the closest you can get is by simplifying.

2x+3x-1+4x+5

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Answer:

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Step-by-step explanation:

For 1, y intercept/x intercept is not 0 when you graph, thus not proportional. Number 2 is, thus proportional. For 3, 4, 5, and 6 you would plot them accordingly. So for (-5, -2) Start as 0,0 and go 5 to the left as it's negative, and 2 down as it is also negative. Do the same for others.

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
Given radical a and radical b are in simplest radical form and radical a timed radical b equals c radical d and c radical d is i
lawyer [7]

Answer:

See below.

Step-by-step explanation:

Since sqrt(a) and sqrt(b) are in simplest radical form, that means a and b have no perfect square factors. When sqrt(a) and sqrt(b) are multiplied giving c * sqrt(d), the fact that c came out of the root means that there was c^2 inside the product sqrt(ab). This means that a and b have at least one common factor.

ab = c^2d

Example:

Let a = 6 and let b = 10.

sqrt(6) and sqrt(10) are in simplest radical form.

Now we multiply the radicals.

sqrt(a) * sqrt(b) = sqrt(6) * sqrt(10) = sqrt(60) = sqrt(4 * 15) = 2sqrt(15)

We have c = 2 and d = 15.

ab = c^2d

6 * 10 = 2^2 * 15

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Our relationship between a, b and c, d works.

6 0
2 years ago
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