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2 years ago

I need the answer please help

Mathematics

1 answer:

olya-2409 [2.1K]2 years ago

5 0

Attached solution with detailed work.

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11m −4/3=2m+5/3<br><br><br><br>pls tell me this answer is the answer 3

Oksana_A [137]

Answer:

m=1/3

Step-by-step explanation:

11m- 4/3=2m+5/3

-2m. - 2m

9m -4/3=5/3

+4/3.+ 4/3

9m= 9/3

m=1/3

6 0

2 years ago

Simplify: −√16x^2 , when x<0

Savatey [412]

Answer:

$-4|x|$

Step-by-step explanation:

To simplify :

$-\sqrt{16x^2}$ , when $x$

In order to simplify the given expression we will take the square root of the term inside the root.

The term in the square root are perfect squares (square of an integer), so, we can write them in square of the numbers.

⇒ $-\sqrt{(\pm4x)^2}$ [As $(\pm4x)^2=16x^2$ ]

On taking square root the square gets removed. So, we have,

⇒ $-(\pm4x)$

⇒ $-4|x|$

Since $x$ , so the value would be negative.

So, simplified answer is

$-4|x|$ (Answer)

3 0

3 years ago

A runner can run 20 miles in 150 mins if she runs at that speed what is her pace in mins per mile

juin [17]

Answer:7.5

Step-by-step explanation:

4 0

2 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

2 years ago

I have $5$ different mathematics textbooks and $4$ different psychology textbooks. In how many ways can I place the $9$ textbook

leonid [27]

Answer:

What

Step-by-step explanation:

You can’t have more than 2 combinations with £9

6 0

2 years ago