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4 years ago

What is the value of x??

Mathematics

1 answer:

nadezda [96]4 years ago

3 0

Answer:

C) square root of 12

Step-by-step explanation:

A^2+b^2=c^2

A=2

B=?

C=4

2^2=4

4^2=16

4+b^2=16

Subtract 4

B^2=12

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3 years ago

Solve the system of equations.

Orlov [11]

Answer:

x=-1, y = 2, z = 1

Step-by-step explanation:

We are given with three equations and we are asked to find the solution to them.

2x + 2y + 3z = 5 ------------- (A)

6x + 3y + 6z = 6 --------------(B)

3x + 4y + 4z = 9 ---------------(C)

Step 1 .

multiplying equation (A) by 3 and subtracting B from the result

6x + 6y + 9z = 15

6x + 3y + 6z = 6

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Step 2.

Substituting this value of y in equation B and C

6x + 3(3-z) + 6z = 6

6x+9-3z+6z=6

6x+3z=-3

2x+z=-1 ----------------(D)

3x + 4(3-z) + 4z = 9

3x+12-4z+4z=9

3x=-3

x=-1 ------------ (E)

Putting this value f x in (D)

2(-1)+z=-1

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y=3-z

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6 0

4 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify

irina [24]

Answer:

x-coordinates of relative extrema = $\frac{-1}{2}$

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

$f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}$

Differentiate with respect to x

$f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}$

$f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}$

Differentiate f'(x) with respect to x

$f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1$

At x = $\frac{-1}{2}$ ,

$f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0$

We know that if $f''(a)>0$ then x = a is a point of minima.

So, $x=\frac{-1}{2}$ is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

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