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3 years ago

30 divided by 2438 long division

Mathematics

2 answers:

meriva3 years ago

7 0

Answer-

81.3
Hope that helped :)

Ksju [112]3 years ago

6 0

The answer would be 81.3

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Meredith has $630,000 she wants to save. If the FDIC insurance limit per

Viefleur [7K]

Answer:

$200,000 in bank A, $200,000 in bank B, $230,000 in bank C

Step-by-step explanation:

Meredith has $630,000

Limit per depositor, per bank, is $250,000

She needs to distribute her money between three Banks to guarantee that her money is insured.

A. $200,000 in bank A, $200,000 in bank B, $230,000 in bank C

It can be seen in A that her deposit per bank deposit didn't exceed the $250,000 limit in the three Banks.

B. $200,000 in bank A, $170,000 in bank B, $260,000 in bank C

Here, her deposit in bank C exceeds $250,000, so there is no guarantee for insurance in bank C

C. $160,000 in bank A, $180,000 in bank B, $290,000 in bank C

Her deposit in bank C is $290,000 which exceeds the $250,000 limit. Therefore, no guarantee for insurance of her money in bank C

D. $160,000 in bank A, $200,000 in bank B, $270,000 in bank C

you

Also, her deposit in bank C exceeds $250,000, so there is no guarantee for insurance in bank C

The way her money can be distributed between three Banks and guarantee insurance is

A. $200,000 in bank A, $200,000 in bank B, $230,000 in bank C.

That way, her deposit per bank is less than the $250,000 limit

3 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

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