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I am Lyosha [343]
3 years ago
15

Try this hard Math Problem if you dare!!

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
6 0

Answer:

a. (x - 3)^2 + 16

b. 8(x -7)^2

c. (a^2 - 1)(7x - 6) or (a+1)(a-1)(7x-6)

d. (x^2-4)(x^2+3) or (x-2)(x+2)(x^2+3)

e. (a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})

Step-by-step explanation:

a.\ (x + 1)^2 - 8(x - 1) + 16

Expand

(x + 1)(x + 1) - 8(x - 1) + 16

Open brackets

x^2 + x + x + 1 - 8x + 8 + 16

x^2 + 2x + 1 - 8x + 24

Collect Like Terms

x^2 + 2x - 8x+ 1  + 24

x^2 - 6x+ 25

Express 25 as 9 + 16

x^2 - 6x+ 9 + 16

Factorize:

x^2 - 3x - 3x + 9 + 16

x(x -3)-3(x - 3) + 16

(x - 3)(x - 3) + 16

(x - 3)^2 + 16

b.\ 8(x - 3)^2 - 64(x-3) + 128

Expand

8(x - 3)(x - 3) - 64(x-3) + 128

8(x^2 - 6x+ 9) - 64(x-3) + 128

Open Brackets

8x^2 - 48x+ 72 - 64x+192 + 128

Collect Like Terms

8x^2 - 48x - 64x+192 + 128+ 72

8x^2 -112x+392

Factorize

8(x^2 -14x+49)

Expand the expression in bracket

8(x^2 -7x-7x+49)

Factorize:

8(x(x -7)-7(x-7))

8((x -7)(x-7))

8(x -7)^2

c.\ 7a^2x - 6a^2 - 7x + 6

Factorize

a^2(7x - 6) -1( 7x - 6)

(a^2 - 1)(7x - 6)

The answer can be in this form of further expanded as follows:

(a^2 - 1^2)(7x - 6)

Apply difference of two squares

(a+1)(a-1)(7x-6)

d.\ x^4 - x^2 - 12

Express x^4 as x^2

(x^2)^2 - x^2 - 12

Expand

(x^2)^2 +3x^2- 4x^2 - 12

x^2(x^2+3) -4(x^2+3)

(x^2-4)(x^2+3)

The answer can be in this form of further expanded as follows:

(x^2-2^2)(x^2+3)

Apply difference of two squares

(x-2)(x+2)(x^2+3)

e.\ a^{4n} -b^{4n}

Represent as squares

(a^{2n})^2 -(b^{2n})^2

Apply difference of two squares

(a^{2n} -b^{2n})(a^{2n} +b^{2n})

Represent as squares

((a^{n})^2 -(b^{n})^2)(a^{2n} +b^{2n})

Apply difference of two squares

(a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})

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Find an equation of the tangent plane to the given parametric surface at the specified point.
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Answer:

Equation of tangent plane to given parametric equation is:

\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

Step-by-step explanation:

Given equation

      r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}---(1)

Normal vector  tangent to plane is:

\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}

\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}

Normal vector  tangent to plane is given by:

r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]

Expanding with first row

\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\

at u=5, v =π/3

                  =\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k} ---(2)

at u=5, v =π/3 (1) becomes,

                 r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

From above eq coordinates of r₀ can be found as:

            r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})

From (2) coordinates of normal vector can be found as

            n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)  

Equation of tangent line can be found as:

  (\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

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Which expression is equivalent to RootIndex 3 StartRoot 32 x Superscript 8 Baseline y Superscript 10 Baseline EndRoot?
evablogger [386]

Answer:

The equivalent expression to the given expression is \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

Step-by-step explanation:

The given expression is \sqrt[3]{32x^8y^{10}}

To find the equivalent expression:

\sqrt[3]{32x^8y^{10}}=(32x^8y^{10})^{\frac{1}{3}}

We may write the above expression as below:

=(32^{\frac{1}{3}})((x^8)^{\frac{1}{3}})((y^{10})^{\frac{1}{3}})

=(2)((4)^{\frac{1}{3}})(x^6)\times (x^2)(x^{\frac{2}{3}})(y^3)(y^{\frac{1}{3}}) (using square root properties)

=(2\sqrt[3]{4})(x^2\sqrt[3]{x^2})(y^3\sqrt[3]{y}) (combining the like terms and doing multiplication )

=2x^2y^3\sqrt[3]{4x^2y}

Therefore \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

Therefore the equivalent expression to the given expression is \sqrt[3]{32x^8y^{10}}=2x^2y^3\sqrt[3]{4x^2y}

5 0
3 years ago
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Answer:

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Step-by-step explanation:

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