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dexar [7]
3 years ago
7

What is the slope of a line that is perpendicular to the line represented by the equation 2x−5=−y?

Mathematics
1 answer:
Liula [17]3 years ago
3 0

Answer:

\frac{1}{2}

Step-by-step explanation:

First, you need to get the equation into y = mx + b form, like this:

Add y to both sides to get y + 2x − 5= 0

Then subtract 2x to get y - 5 = -2x

Add 5 to get y = -2x + 5

Now we know the slope is -2, or -2/1.

A line that is perpendicular to it has a slope that is the first line's slope's opposite reciprocal. This means flip the fraction upside down and change the sign, which gives you positive 1/2.

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Using the Factor Theorem, which of the polynomial functions has the zeros 2, radical 3 , and negative radical 3 ? f (x) = x3 – 2
butalik [34]

To solve this question, we use the factor theorem, and using it, the polynomial function is:

f(x) = x^3 - 2x^2 - 3x + 6

------------------------------

The factor theorem means that if k is a root of f(x), f(k) = 0.

Thus, applying the factor theorem for this question, we have to choose the function for which: f(2) = 0, f(\sqrt{3}) = 0, f(-\sqrt{3}) = 0

------------------------------

Function 1:

f(x) = x^3 - 2x^2 - 3x + 6

Testing the values:

f(2) = 2^3 - 2(2)^2 - 3(2) + 6 = 8 - 8 - 6 + 6 = 0

f(\sqrt{3}) = \sqrt{3^3} - 2(\sqrt{3})^2 - 3\sqrt{3} + 6 = \sqrt{3^2\times3} - 6 - 3\sqrt{3} + 6 = 3\sqrt{3} - 3\sqrt{3} = 0

f(-\sqrt{3}) = -\sqrt{3^3} - 2(-\sqrt{3})^2 - 3(-\sqrt{3}) + 6 = -\sqrt{3^2\times3} - 6 + 3\sqrt{3} + 6 = -3\sqrt{3} + 3\sqrt{3} = 0

Thus, since all three conditions are satisfied, f(x) = x^3 - 2x^2 - 3x + 6 is the polynomial function.

A similar question is given at brainly.com/question/11378552

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If line A and B are parallel, then both angle 6 and 8 are corresponding angles, which mean they are the same.

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Step-by-step explanation:

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