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3 years ago

Of the following points, name all that lie on the same horizontal line? (1,-2) (-1,2) (-2,1) (2,-1)

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

8 0

Answer:

None

Step-by-step explanation:

Let’s graph the points first.

Look at the image below.

And by looking at the image we can tell that none of the points lay on the same horizontal line.

Thus,

none of the points lay on the same horizontal line.

Hope this helps :)

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Points E, D, and H are the midpoints of the sides of ΔTUV. UV=56, TV=72, and HD=56. Find HE.

natita [175]

We need a diagram to do this

5 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

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

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

Find the common fraction equivalent to 0.12

jonny [76]

The answer is: " $\frac{4}{33}$ "  .
______________________________
Given: 0.1212121212..... repeating ; write that value as a fraction;
______________________________________________________
In other words; we are given: "0.1212121212..... repeating infinitely" ;

→ that is to say; "0.12 ... ; {the "12" decimal portion repeats infinitely} ;
_______________________________________________________
→ We write this value, as a fraction, as: "12/99" ;
__________________________________________
→Explanation:
__________________________________________

Note: "0.99999999...... repeating infinitely; = "1" .
_________________________________________
→Since:

Let us say that we have:

"x = 0.999999 ; repeating infinitely;

In order words, let us say we have: "x = 0.9.... ; the "9" decimal repeats infinitely;
_____________________________________________________
Then "10x" ; (that is: "10" multlipled by "x"; or "10*x" or "10x" ); is equal to:

"10" * (0.999999.....) = 9.99999999...... (the "9" decimal repeats infinitely);

in other words: 10x = 9.99999999....

Divide each side by "10" ;

to get "x = 0.9999999....." ; the decimal "9" repeats indefinitely...." ;

But if you have: "10x = 10" ; divide each side of the equation by "10" ;
you get: "x = 1" .
____________________
Also, if "x = 0.9999...(repeating infinitely);

then: 10x = 9.99999.
_______________________________________________
10x = 9.999999999999999......
−   x = 0.999999999999999.......
_____________________________________
9x = 9.00000000000000000000.....

→ 9x = 9 ;

Divide each side of the equation; to get;

9x /9 = 9/ 9 ; to get: x = 1 ; and we have: x = 0.9999.... ; so
x= 0.99999.... = 1 ;
__________________________________________________
So, if the numbers "12" is repeating, we divie "12" by "99" ;
that is; we divide "12" by "two 9's" ; since "12" is a "TWO-digit number"; a "two-digit number" is being repeated infinitely.
________________________________________
So; 0.12121212.....(the "12" is the decimal that repeats infinitely);

= 12/99 ; which can be simplified;

Divide each side (both the numerator AND the denominator); by "3" ;
_________________________________________
" 12/99 " = "(12÷3) / (99÷3) = 4/33 " .
_________________________________________
The answer is: $\frac{4}{33}$   .
_________________________________________

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3 years ago

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