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2 years ago

Find the quotient 4/3÷1 1/12

Mathematics

1 answer:

nlexa [21]2 years ago

8 0

The answer should be 1/9

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The drawing of a building shown below has a scale of 1 inch to 30 feet. What is the acual height in feet of the building?

mylen [45]

Multiply the amount of inches it is by 30 and you will know what the size is in feet.

8 0

3 years ago

Read 2 more answers

Consider the 20 points of a 4×5 grid. You randomly choose two points from the 20 points. What is the probability that the two po

Serhud [2]

Answer:

The probability of both points falling in the same row or column is 7/19, or approximately 37%

Step-by-step explanation:

The easiest way to solve this is to think of it rephrased as "what is the probability that your second point will be in the same row or column as your first point". With that frame of reference, you can simply consider how many other points are left that do or do not fall in line with the selected one.

After selecting one, there are 19 points left.

The row that the first one falls in will have 3 remaining empty points.

The column will have 4 remaining empty points.

Add those up and you have 7 possible points that meet the conditions being checked.

So the probability of both points falling in the same row or column is 7/19, or approximately 37%

6 0

2 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

Please Help ASAP! I suck at geometry :(

ZanzabumX [31]

We know that for every 5 red bricks there were 2 gray bricks.

The total amount of red bricks and grey bricks in this sample is 7.

5 red bricks + 2 grey bricks = 7 bricks

Now, we divide 175 "total number of bricks used" by 7 "5 red bricks + 2 grey bricks = 7 bricks" and we will get a quotient of 25.

Now we know that 25 bricks is $\frac{1}{7}$ of the wall. The gray bricks are $\frac{2}{7}$ so we can multiply 25 by 2 and we will get a product of 50. If 1/7 = 25 grey bricks so 2/7 would be the grey bricks.