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ladessa [460]
3 years ago
6

Please help me! I'm a bit skeptical about my answer, is it correct or not?

Mathematics
1 answer:
lana66690 [7]3 years ago
3 0
Yea the answer you chose is right!
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What is the derivative of e to the poker minus x
WARRIOR [948]
(e^{-x})'=e^{-x}\cdot(-x)'=e^{-x}\cdot(-1)=-e^{-x}
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Do you know the answer
aivan3 [116]

Answer:

84.3

Step-by-step explanation:

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Identify the zeros of the function f(x) = 2x^2 − 4x + 5 using the Quadratic Formula
Nostrana [21]

For this case we have that by definition, the roots, or also called zeros, of the quadratic function are those values of x for which the expression is 0.

Then, we must find the roots of:

2x ^ 2-4x + 5 = 0

Where:

a = 2\\b = -4\\c = 5

We have to:x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

Substituting we have:

x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (5)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-40}} {4}\\x = \frac {4 \pm \sqrt {-24}} {4}

By definition we have to:

i ^ 2 = -1

So:

x = \frac {4 \pm \sqrt {24i ^ 2}} {4}\\x = \frac {4 \pm i \sqrt {24}} {4}\\x = \frac {4 \pm i \sqrt {2 ^ 2 * 6}} {4}\\x = \frac {4 \pm 2i \sqrt {6}} {4}\\x = \frac {2 \pm i \sqrt {6}} {2}

Thus, we have two roots:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

Answer:

x_ {1} = \frac {2 + i \sqrt {6}} {2}\\x_ {2} = \frac {2-i \sqrt {6}} {2}

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lora16 [44]

Answer:

False, A equation DOES have to have a equal sign.

Step-by-step explanation:

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Rewrite the equation y - 4= 2(x + 4) in slope-intercept form.
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First distribute the 2 through the parenthses on the right side.

So we have y - 4 = 2x + 8.

In slope-intercept form, the y is by itself on the left side.

So we add 4 to isolate the y on the left to get y = 2x + 12.

So in slope-intercept form, our equation is y = 2x + 12.

7 0
3 years ago
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