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3 years ago

Consider the function f(x)=-5x^2 +2x -8. find the critical point a of the function

1 answer:

5 0

To find the critical number of that function we have to find the derivative and set it equal to 0 and solve for x. The derivative of the function is f'(x)=-10x+2. If we set it equal to 0 and solve, we have x = -2/-10 which simplifies to x = 1/5.

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Brainliest!!! CORRECT and fast answers. :)

Vladimir79 [104]

Answer:

A. 2 units left and 3 units up

Step-by-step explanation:

2 from x is taken away and thus shifts 2 to the left

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3 years ago

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crimeas [40]

Answer: number 1

Step-by-step explanation:

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3 years ago

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natta225 [31]

Answer:

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3 years ago

Find [4(cos15° + isin15°)]^3 and express it in a + bi form.

arsen [322]

Answer:

see explanation

Step-by-step explanation:

Using De Moivre's theorem

Given

[ 4(cos15° + isin15° ) ]³, then

= 4³ [ cos(3 × 15°) + isin(3 × 15°) ]

= 64 (cos45° + isin45° )

= 64 ( $\frac{\sqrt{2} }{2}$ + $\frac{\sqrt{2} }{2}$ i )

= 64 ( $\frac{\sqrt{2} }{2}$ (1 + i) )

= 32 $\sqrt{2}$ (1 + i)

= 32 $\sqrt{2}$ + 32 $\sqrt{2}$ i

3 0

3 years ago

Solve (x + 1)2 – 4(x + 1) + 2 = 0 using substitution.

solniwko [45]

For this case we have to:

Let $u = x + 1$

So:

$u ^ 2-4u + 2 = 0$

We have the solution will be given by:

$u = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Where:

$a = 1\\b = -4\\c = 2$

Substituting:

$u = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (1) (2)}} {2 (1)}\\u = \frac {4 \pm \sqrt {16-8}} {2}\\u = \frac {4 \pm \sqrt {8}} {2}\\u = \frac {4 \pm \sqrt {2 ^ 2 * 2}} {2}\\u = \frac {4 \pm2 \sqrt {2}} {2}$

The solutions are:

$u_ {1} = \frac {4 + 2 \sqrt {2}} {2} = 2 + \sqrt {2}\\u_ {2} = \frac {4-2 \sqrt {2}} {2} = 2- \sqrt {2}$

Returning the change:

$2+ \sqrt {2} = x_ {1} +1\\x_ {1} = 1 + \sqrt {2}\\2- \sqrt {2} = x_ {2} +1\\x_ {2} = 1- \sqrt {2}$

Answer:

$x_ {1} = 1 + \sqrt {2}\\x_ {2} = 1- \sqrt {2}$

3 0

3 years ago

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