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3 years ago

If f(x)=x+6 and g(x)=x^4 find G(F(x))

Mathematics

2 answers:

lyudmila [28]3 years ago

7 0

Answer:

$\large\boxed{g(f(x))=(x+6)^4}$

Step-by-step explanation:

$f(x)=x+6\\\\g(x)=x^4\\\\g(f(x))\to\text{put}\ x+6\ \text{to the equation of}\ g(x)\\\\g(f(x))=(x+6)^4$

Novay_Z [31]3 years ago

3 0

You plug in f(x) into where x is within g(x) so you get G(f(x))=(X+6)^4

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

Neko [114]

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:

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

5 0

3 years ago

Find the distance between (4,-2) & (-1,3)

AleksAgata [21]

Step-by-step explanation:

Use Pythagorean theorem here. To find the height, find the difference of the 2 y's. To find the length, find the difference between the 2 x's.

Length = x2 - x1

Height = y2 - y1

√length^2 + height^2 = distance

√5^2 + 5^2 = √50 = 7.07

8 0

3 years ago

The total area of the figure below is (blank) units look at picture to solve

Tpy6a [65]

9x4=36 squared units

3 0

3 years ago

What is the length of MN?

alexira [117]

Answer:

MN = 6

This is because the shapes are dialations of each other

So PO and ML are the same side of the triangles but ML is dialated by a factor of 3.5 so that applies to the whole shape. So 21/3.5 = 6

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2 years ago

Dana and Steven read the same book with 60 pages. Dana read 1/4 of the book in 3/10 hour. Steven read 1/5 of the book in 1/6 hou

goblinko [34]

Answer: Steven read more pages per minute than Dana

Step-by-step explanation:

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3 years ago