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3 years ago

Help????? urgent!!!!!!!!

Mathematics

1 answer:

defon3 years ago

3 0

Y=2x-3.
Answer B.

Proof

When y=1 then x=2
y=2x-3
1=2*2-3
1=4-3
1=1

.............

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-(1/x+3) +4

nlexa [21]

9514 1404 393

Answer:

left 3 units
up 4 units
shape: lower left image

Step-by-step explanation:

For a parent function f(x), the transformations ...

g(x) = a×f(x -h) +k

cause ...

vertical expansion by 'a', reflection over x-axis if negative
right shift by 'h'
up shift by 'k'

Here, we have parent function f(x) = 1/x with a=-1, h=-3, k=4. Then the transformations are ...

horizontal shift left 3 units

vertical shift up 4 units

reflection over x-axis, so curves are above-left and below-right of the reference point (Note that the reflection is done <em>before</em> the translation.)

4 0

2 years ago

What is the ratio for 45 miles and 0.75 hours

alisha [4.7K]

Answer:

I'm pretty sure it is 60 miles per hour (60:1)

Step-by-step explanation:

I used a unit rate calculator

Sorry if I'm wrong

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3 years ago

Simplify √18 ÷ √12 please help explain this

qaws [65]

√18 ÷ √12 = 1.22474487139159

8 0

3 years ago

Read 2 more answers

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

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