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3 years ago

Find the measure of one interior angle in a regular 144-gon. Round to the nearest

Mathematics

1 answer:

lubasha [3.4K]3 years ago

8 0

Answer:

the answer is d

Step-by-step explanation:

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EP6. Abby works at a job that starts out paying $15 per hour. Each year she continues

Artemon [7]

Answer:

$64.83.

Step-by-step explanation:

Each year the multiplier fffor her new rate will be 1.05.

So after 30 years her hourly rate will be 15 * 1.05^30

= $64.83

6 0

2 years ago

Write A expression for increased by 23

Akimi4 [234]

Answer:

5+23

Step-by-step explanation:

increased by is another word for add

7 0

3 years ago

Read 2 more answers

The largest butterfly in the world is found Papua New Guinea.the female of the species weights about 9/10 Use a decimal to write

Anarel [89]

Well, 9/10 is the same as 9 10ths, the decimal that represents that is 0.9 because it is also 9 10ths

5 0

3 years ago

A dome of a building is in the form of a hemisphere from inside it was whitewashed at the cost of rupees 4989.60 if the cost of

Luda [366]

Answer: $249.48\ m^2; 369.168\ m^3$

Step-by-step explanation:

Given

Cost of white-washing =Rs. 4989.60

Rate =20 per meter square

So, inside area of = $\dfrac{4989.60}{20}=249.48\ m^2$

The surface area of dome= $4\pi\ r^2=249.48$

$4\times 3.142\times r^2=249.48\\r^2=\frac{249.48}{4\times 3.142}=19.85\\r=4.455\ m$

The volume of air in the dome $V=\frac{4}{3}\pi r^3$

$V=\frac{4}{3}\cdot 3.142\cdot (4.45)^3\\V=369.17\ m^3$

6 0

3 years ago

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