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3 years ago

Please help!!! I’m going to insert bs equation here 4 + 6 = 10 is an example of an equation.

Mathematics

1 answer:

Cloud [144]3 years ago

6 0

Answer:

heart exclamation mark

Step-by-step explanation:

you could also maybe make it into a really badly done pokeball

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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

Please please help im confused again

timama [110]

Answer:

$A_s=179 \ yd$

Step-by-step explanation:

Area of Circle can be calculated using following formula

$A=\pi r^{2}$

Where r is radius of circle

<u>Area of outer circle (r=11 yd)</u>

$A_1=\pi (11)^{2}\\A_1=379.94 \ yd$

<u>Area of inner circle (r=8 yd)</u>

$A_2=\pi (8)^{2}\\A_2=200.96 \ yd$

Area of shaded region

$A_s=A_1-A_2$

$A_s=379.94-200.96\\A_s=179 \ yd$

3 0

3 years ago

A runner has run 1.192 kilometers of a 10-kilometer race. How much farther does he need to run to finish the race?

Marianna [84]

Answer is 8.08 kilometers

4 0

2 years ago

Read 2 more answers

Ayuda es urgente ya lalalalalalalalalalal

Crank

Answer:

no se ve la imagen perdón no te puedo ayudar

8 0

3 years ago

each year francesca earns a salary of 2% higher than her previous years salary. In her first 5 years at this job, she earned a t

Luda [366]

Answer:

$36,027.9

Step-by-step explanation:

Lets say her salary in the first year is s1. So, in the second year her salary will be 2% higher than x, this is, s1 + s1*0.02, so:

s2 = s1 + 0.02s1 = 1.02s1

Then in year 3 it will be s2 plus a 0.02 increase:

s3 = s2 + 0.02s2 = 1.02(s2) = 1.02 (1.02 s1) = 1.02^2(s1)

Then the same for s4: s3 salary plus the increase.

s4 = s3 + 0.02s3 = 1.02(s3) = 1.02 (1.02 s2) = 1.02(1.02*1.02 s1) =1.02^3(s1)

Finally for the 5th monts:

s5 = 1.02(s4) = 1.02^4(s1)

We know that s1+s2+s3+s4+s5 = 187,345

Lets replace all the salaries by its value depending on s1

s1 + 1.02s1 + 1.02^2(s1) + 1.02^3(s1) + 1.02^4(s1) = 187,345

We can add to the first term a s1^0 as it is eqaul to 1

s1 + 1.02s1 + 1.02^2(s1) + 1.02^3(s1) + 1.02^4(s1) = 187,345

Lets get 1s1 out as common factor:

s1 (1 + 1.02^1 + 1.02^2 + 1.02^3 + 1.02^4) = 187,345

s1 *5.20 = 187,345

Divide both sides by 5.20

s1 = 36,027.9

So, her salary in the 1st month was approximately $36,027.9

4 0

2 years ago

Read 2 more answers

Please help!!! I’m going to insert bs equation here *4 + 6 = 10 is an example of an equation.*

Please help!!! I’m going to insert bs equation here 4 + 6 = 10 is an example of an equation.