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Gwar [14]
3 years ago
7

The two tables give the number of pints of red and blue paint that are used to make different amounts of two shades of purple pa

int.
Purple #1 is made by mixing red and blue in a ratio of 1 : 2.

Purple #2 is made by mixing red and blue in a ratio of 1 : 3.


Use the drop-down menus to complete each sentence below.


Math item stem image


The two highlighted rows show that for the same amount of blue, Purple #1 uses (less,more)red than Purple #2.


This means that Purple #1 is(a redder,a bluer, the same) shade of purple than Purple #2.


Purple #2 is(a redder,a bluer, the same) shade of purple than Purple #1.

Mathematics
2 answers:
kirza4 [7]3 years ago
5 0

The two highlighted rows show that for the same amount of blue, Purple #1 uses <u>more</u> red than Purple #2.

This means that Purple #1 is <u>a redder</u> shade of purple than Purple #2.

Purple #2 is <u>a bluer</u> shade of purple than Purple #1.

Step-by-step explanation:

The two highlighted rows show that for the same amount of blue, Purple #1 uses <u>more</u> red than Purple #2.

  • Making blue's quantity as 3 parts for purple #1 implies red part becomes 1.5 to maintain the ratio 1:2
  • Purple #1 has 1/3 parts red and 2/3 parts blue. Purple #2 has 1/4th part red and 3/4th part blue.
  • Hence, Purple #1 is <u>a redder</u> shade of purple than Purple #2.
  • From the above explanation, Purple #2 is <u>a bluer</u> shade of purple than Purple #1.
Cerrena [4.2K]3 years ago
5 0

Answer:

The two highlighted rows show that for the same amount of blue, Purple #1 uses <u>More</u> red than Purple #2.

This means that Purple #1 is <u>Redder</u> shade of purple than Purple #2.

Purple #2 is <u>Bluer</u> shade of purple than Purple #1.

Step-by-step explanation:

Because I had to do it too

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Answer:

Equation of tangent plane to given parametric equation is:

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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})

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