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3 years ago

Which number has an absolute value greater than 4?

Mathematics

1 answer:

pav-90 [236]3 years ago

5 0

Answer:

Step-by-step explanation:

Sorry if this doesn't help much and if its wrong, but maybe you could mark this as brainly? Have a wonderful day!

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If the length of the base of a triangle is

Anarel [89]

Answer:

Step-by-step explanation:

the answer is 12% decrease in area

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3 years ago

The Marshall Island goby, the world's smallest fish, measures .47 inch. The striped bass, an American sport fish, is about 25.5

OLEGan [10]

150 inches because i say it is because i just need the points to ask another question

6 0

3 years ago

Read 2 more answers

interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm

Igoryamba

Answer:

The curvature is $\kappa=1$

The tangential component of acceleration is $a_{\boldsymbol{T}}=0$

The normal component of acceleration is $a_{\boldsymbol{N}}=1 (2)^2=4$

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}$

where

$\boldsymbol{T}}$ is the unit tangent vector.

$\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ is the speed of the object

We need to find $\boldsymbol{r}'(t)$ , we know that $\boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k}$ so

$\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}$

Next , we find the magnitude of derivative of the position vector

$|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2$

The unit tangent vector is defined by

$\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}$

$\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)$

We need to find the derivative of unit tangent vector

$\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})$

And the magnitude of the derivative of unit tangent vector is

$||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2$

The curvature is

$\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1$

The tangential component of acceleration is given by the formula

$a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}$

We know that $\frac{ds}{dt}=|| \boldsymbol{r}'(t)}||$ and $||\boldsymbol{r}'(t)}||=2$

$\frac{d}{dt}\left(2\right)\: = 0$ so

$a_{\boldsymbol{T}}=0$

The normal component of acceleration is given by the formula

$a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2$

We know that $\kappa=1$ and $\frac{ds}{dt}=2$ so

$a_{\boldsymbol{N}}=1 (2)^2=4$

3 0

3 years ago

A LARGE PIZZA AT TONY'S PIZZERIA IS A CIRCLE WITH

AfilCa [17]

You already know what it is cuh

5 0

3 years ago

Read 2 more answers

A Norman window is a window with a semi-circle on top of regular rectangular window. (See the picture.) What should be the dimen

Vikki [24]

Answer:

bottom side (a) = 3.36 ft

lateral side (b) = 4.68 ft

Step-by-step explanation:

We have to maximize the area of the window, subject to a constraint in the perimeter of the window.

If we defined a as the bottom side, and b as the lateral side, we have the area defined as:

$A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}$

The restriction is that the perimeter have to be 12 ft at most:

$P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12$

We can express b in function of a as:

$2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a$

Then, the area become:

$A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2$

To maximize the area, we derive and equal to zero:

$\dfrac{dA}{da}=6-2\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right )a=0\\\\\\6-(1-\pi/4)a=0\\\\a=\dfrac{6}{(1+\pi/4)}\approx6/1.78\approx 3.36$

Then, b is:

$b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a\\\\\\b=6-0.393*3.36=6-1.32\\\\b=4.68$

3 0

3 years ago