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3 years ago

Which statement can you use to conclude that RSTW is a parallelogram

Mathematics

1 answer:

Romashka [77]3 years ago

3 0

It is answer choice a

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Use the law of cosines to find the length of a

borishaifa [10]

Answer:

Step-by-step explanation:

You don't need the Law of Cosines, the Law of sines if what you need. You can't use the Law of Cosines because in order to find side a, you would need the length of side c and you don't have it. Using the Law of Sines is appropriate, knowing that angle B = 55:

$\frac{sin55}{175}=\frac{sin42}{a}$ and solving for a:

$a=\frac{175sin42}{sin55}$ so

a = 143.0

4 0

3 years ago

Answer my math question I asked so many times I lost most of points by

kumpel [21]

Answer:

i think ur in a higher grade then me but im good w/ equations so OoOO0p

Step-by-step explanation:

equation: y = x - 6 + x^2

(6,0) (7,1) (8,4) (9,9)

I figured this out by looking at the point (6,0). To get 0 (y), you have to subtract 6. Knowing this, I subtracted 6 from the rest of the coordinates, leaving me with numbers that are able to be squared to get y. This led me to the equation x - 6 + x^2.

8 0

3 years ago

Which integer represents a golf score of 4 over par?

fgiga [73]

Omg ur learning about integers, I'm learning about integers too, what grade are you in?

7 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

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