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podryga [215]
3 years ago
8

Step by step please. im stuck .

Mathematics
1 answer:
scoundrel [369]3 years ago
7 0

Answer:

2u³

Step-by-step explanation:

\sqrt[7]{128w^{21}} =  \sqrt[7]{2^{7}(w^{3})^{7}} =2u^{3}

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Which of the following is a proportional relationship?
Solnce55 [7]

The two equations represent the proportional relationship.

y=3x and y=12x are proportional relation ship equations

proportion equations can be defined as

If we change x the y will change in the same proportion.

<h3>What is the proportional relationship?</h3>

Proportional relationships are relationships between two variables where their ratios are equivalent.

Another way to think about them is that, in a proportional relationship, one variable is always a constant value time the other.

That constant is known as the constant of proportionality.

proportional relationship equation contain (0,0) points

If we put x=0

This  will give us,y=0

If we put x=0, in y=12x

It will give y=0

put if we put x=0 in

y=3x it will give us y=0

hence these two equations represent the proportional relationship.

To learn more about the equation visit:

brainly.com/question/2972832

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6 0
2 years ago
A shoreline is eroding at a rate of 1/3 3 foot per 3 months. How many feet are eroding per year?
salantis [7]
7ft if 3+1/3 hopefully this helps i did hwat i can










4 0
4 years ago
Read 2 more answers
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
3x = 8x – 5 es para mañana
Volgvan

Answer:

x=1

Step-by-step explanation:

3x=8x-5

-8x -8x

-5x=-5

divide by -5 on each side

x=1

5 0
3 years ago
Read 2 more answers
I need to find mGM ​
Rus_ich [418]

Answer:

if you're still looking at it is not the problem for you have the time of me and the other day that we will not for sure if I

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