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aleksandrvk [35]
3 years ago
9

What does the discriminant tell us if it is -10? I need help by the end of today!

Mathematics
1 answer:
amid [387]3 years ago
7 0

Answer:

theres no picture that comes with it?

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Algebra 2 Questions:
Anvisha [2.4K]
The answer to 1 is (E) All of the above. This is because a negative cannot be under a square root and it comes out as an i. 

The answer to 2 is (B) a is real and b is imaginary. This is because anything that has an "i" attached to it is considered imaginary.
3 0
3 years ago
What ratio is equivalent to 8 to 2
Luba_88 [7]

Answer:

16 to 4

Step-by-step explanation:

8 to 2 can be reduced to 4, so to have a ratio equivalent, it must be able to be reduced to 4 also. An example would include, 16 to 4.

3 0
3 years ago
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Anybody know how to solve this?
avanturin [10]
The answer would be A. I just did this yesterday. Do you want the steps?

4 0
3 years ago
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Select the correct answer.
kherson [118]

Answer:

d.2/6 has a repeating decimal form

3 0
3 years ago
Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?
Pepsi [2]

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

f(x) = \dfrac{2}{x - 1} + 4

From the given function, we have;

When x = 1, the denominator of the fraction, \dfrac{2}{x - 1}, which is (x - 1) = 0, and the function becomes, \dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as <em>x</em> increases, the fraction \dfrac{2}{x - 1} tends to 0, therefore as x increases, we have;

\lim_  {x \to \infty}  \dfrac{2}{(x - 1)} \to 0, and \  \dfrac{2}{(x - 1)}  + 4 \to 4

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, \dfrac{2}{0 - 1} + 4 = 2. When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

5 0
2 years ago
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