All categories

2 years ago

How do you solve this? -5√112

Mathematics

1 answer:

3241004551 [841]2 years ago

3 0

Simplify the radicand by breaking the radicand up into a product of known factors.

Answer: -20<span>√7

</span>

You might be interested in

What is the 100th term of 6,10,14,18

Usimov [2.4K]

Answer:

406

Step-by-step explanation:

6+(4x100)

3 0

2 years ago

Amy cut a piece of yarn 3.05 inches long for a bracelet she was making. Which length is equivalent to 3.05 inches?

Lesechka [4]

Answer:

C. 3 1/20

Step-by-step explanation:

3.05 as a fraction = 3 1/20

Therefore the answer is C.

7 0

2 years ago

Read 2 more answers

Which is the graph of the linear equation X=-2

tankabanditka [31]

Answer:

photomath thats what u should use or put it in a calculator

3 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

2 years ago

Nastasia [14]

Hey there! I'm happy to help!

Lines that are parallel have the same slopes because they are increasing or decreasing at the same rate and therefore will never bump into each other.

We see that y=2 has a slope of 0 because there is no x that we can see in the equation. This is just a flat line.

If the line we are looking for is flat; it stays at the same y-value the entire time. We see that the y-value for this line of ours is -6. Therefore, the answer is C. y=-6.

I hope that this helps! Have a wonderful day!

3 0

3 years ago