All categories

3 years ago

Can you help on question 22?!

Mathematics

1 answer:

Paladinen [302]3 years ago

7 0

Answer:

I think it may be B (d-$5 )

You might be interested in

Find two consecutive whole numbers that 7 lies between.

Wittaler [7]

The answer is actually 3 and 4.

$\sqrt{7}$ lies between the two consecutive numbers 3 and 4.

I hope this helped! :))

5 0

3 years ago

an architect created a scale model of a building. the length of the model is 15 inches. the scale factor used to create the mode

navik [9.2K]

The length of model is ......15 inch
if the scale factor is 1/20
it means 1/20=15/h
h=12*15=300

4 0

3 years ago

Need help on this one

Advocard [28]

Answer:a

Step-by-step explanation:

That’s one cracked screen

3 0

3 years ago

Read 2 more answers

What is 16/48 equivalent to?

liberstina [14]

$\frac{16}{48}=\\\\\frac{16:2}{48:2}=\boxed{\frac{8}{24}}\\\\\frac{8:2}{24:2}=\boxed{\frac{4}{12}}\\\\\frac{4:2}{12:2}=\boxed{\frac{2}{6}}\\\\\frac{2:2}{6:2}=\boxed{\frac{1}{3}}\\\\\frac{1\cdot10}{3\cdot10}=\boxed{\frac{10}{30}}\\\vdots$

6 0

3 years ago

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

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