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Yuki888 [10]
1 year ago
12

Which of the following points is NOT visible if the viewing window of a graphing utility is set at:

Mathematics
1 answer:
Arte-miy333 [17]1 year ago
3 0

The points D(-7, 5), E(-8, -3), and H(10, -5) will not be visible the answers are D, E, and H.

<h3>What is graphing utility?</h3>

A graphing calculator is a portable computer that can plot graphs, solve multiple equations at once, and carry out other variable-related activities.

We have points shown in the coordinate plane.

As the window of a graphing utility is set at:

X(min) : -6

X(max): 6

Y(min) : -6

Y(max) : 6

The points out of 6 will not be visible.

We can draw a square as shown in the attached picture.

The points D(-7, 5), E(-8, -3), and H(10, -5) will not be visible.

Thus, the points D(-7, 5), E(-8, -3), and H(10, -5) will not be visible the answers are D, E, and H.

Learn more about the graphing utility here:

brainly.com/question/1549068

#SPJ1

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Step-by-step explanation:

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If y=x^2+kx-k for what values of k will the quadratic have 2 real solutions
xeze [42]

The key to answering this question lies in finding the discriminant b^2-4ac. If b^2-4ac is greater than zero, the quadratic equation will have 2 real solutions.

Typing out y=x^2+kx-k, we see that a=1, b=k and c = -k. The discriminant in this case is

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For which k is this k^2 + 4k = 0? Factor k^2 + 4k, obtaining k(k+4) = 0; the roots of this are k=0 and k = -4.

Now set up a number line for k, drawing empty circles at k = 0 and k = -4. These two values divide the number line into three intervals:

(-infinity, -4), (-4, 0) and (0, infinity). Choose a test number for each interval:

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So: The given quadratic will have two different, real roots on the two intervals (-infinity, -4) and (0, infinity).

4 0
3 years ago
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