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3 years ago

Can you help me with this problem I can't find it out

Mathematics

2 answers:

AnnZ [28]3 years ago

8 0

He estimates that he will earn $44,800.

12% of 40,000 is 4,800
40,000+4,800=44,800

zhenek [66]3 years ago

5 0

Money earned last year = $40,000
Money estimated to be earned this year = $40,000 + 12% of $40,000
= $40,000 + (12/100 × $40,000)
= $40,000 + $4,800
= $44,800
Plz mark it brainliest

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Answer:

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Step-by-step explanation:

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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

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

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

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

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3 years ago

Write an equation of the line that passes through a pair of points:

Shalnov [3]

Answer:

Step-by-step explanation:

(-2,-2) , (5,-5)

$Slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{-5-(-2)}{5-(-2)}\\\\=\frac{-5+2}{5+2}\\\\=\frac{-3}{7}\\$

(-2,-2), m= -3/7

Line equation: y -y1 =m*(x -x1)

y-(-2) = -(-3/7)(x - [-2])

$y+2=\frac{-3}{7}*(x+2)\\\\y+2=\frac{-3}{7}*x+(\frac{-3}{7})*2\\\\y+2=\frac{-3}{7}x-\frac{6}{7}\\\\y=\frac{-3}{7}x-\frac{6}{7}-2\\\\y=\frac{-3}{7}x-\frac{6}{7}-\frac{2*7}{1*7}\\\\y=\frac{-3}{7}x-\frac{6}{7}-\frac{14}{7}\\\\y=\frac{-3}{7}x-\frac{20}{7}$

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3 years ago

Read 2 more answers

HURRY PLEASE!!! IM BEING TIMED. I WILL GIVE YOU BRAINLIEST!

mrs_skeptik [129]

Answer:

Step-by-step explanation:

the y intercept three is 4 away from -1

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3 years ago