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3 years ago

Ur answer will be marked as brainliest

Mathematics

1 answer:

Mars2501 [29]3 years ago

7 0

Answer:

Hi there!

The correct answer is: $2x^{2} + y^{2} + z^{2} - 2\sqrt{2} xy - 2\sqrt{2} xz + 2yz$

Step-by-step explanation:

First of all I want to clarify I used to the fourth identity to solve this problem.

The fourth identity is: $(x+y+z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2xz + 2yz$

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What is the axis of symmetry of the quadratic function y=2(x+3)^2+5

Pavel [41]

Answer:

x=-3

Step-by-step explanation:

The quadratic function $y=2(x+3)^2+5$ determines the parabola which has the vertex at point (-3,5). The axis of symmetry passes through the vertex and is parallel to the y-axis. Hence, its equation is x=-3 (see attached diagram).

3 0

3 years ago

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.

$x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0$

$x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0$

$x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0$

$x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0$

The sign of $f''$ changes to either side of $x=1$ and $x=3$ , but not $x=0$ . This means only $\boxed{x=1}$ and $\boxed{x=3}$ are inflection points.

4 0

1 year ago

Read 2 more answers

W many large

ankoles [38]

Answer:

Step-by-step explanation:

40 / 7.50 = 5.3 recurring

Seeing as you can't buy 0.3 of a pizza, we just use the 5

8 0

3 years ago

Read 2 more answers

(05.03)Calculate the area of the regular pentagon below: A regular pentagon with side length of 22.3 inches and dotted line from

Marat540 [252]

Answer:

The area of the regular pentagon is $A=858.55\ in^{2}$

Step-by-step explanation:

we know that

The area of a regular polygon is equal to

$A=\frac{1}{2}Pa$

where

P is the perimeter of the regular polygon

a is the apothem

Find the perimeter P

$P=5(22.3)=111.5\ in$ ----> the regular pentagon has five equal sides

$a=15.4\ in$ ---> distance from center to middle of side

substitute

$A=\frac{1}{2}(111.5)(15.4)$

$A=858.55\ in^{2}$

5 0

3 years ago

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Pls give answer I need to submit today

BabaBlast [244]

Answer:

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7 0

3 years ago