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y=10sin(1π/9*θ+4π0)-17 I need help with finding the period and horizontal translation, whats the formula?

Is this what you intended?    4pi*0 was rather strange = it equals zero.   ://

\(y=10sin(\frac{\pi\theta}{9}+4π*0)-17\\ y=10sin(\frac{\pi}{9}\theta)-17\\ Period=2\pi\div \frac{\pi}{9}\\ Period=2\pi\times \frac{9}{\pi}\\ Period=18\\ Amplitude =10\\ \mbox{horizontal translation is 17 units down } \)

Here is the graph

Thanks, Melody.....I just graphed these in WolframAlpha and took a  "screen snip"

Thanks Chris,

What graphing program did you use Chris?

I played with this question too and I found it quite interesting. My results were the same as yours ofcourse.  :)

Here are the Desmos graphs, 

https://www.desmos.com/calculator/zcpydgu9sb

\(If \quad x=t^2\;\quad y=t+1\qquad \mbox{The restriction is } x\ge0\\ t=\pm\sqrt{x}\qquad t=y-1\\ y-1=\pm \sqrt{x}\\ (y-1)^2=x\\ x=(y-1)^2\\ \)

As Chris said this is a sidways parabola opening in the positive direction with a vertex at (0,1)

Lets look at the second one

a) x=sin^2(t), y=sin(t)+1

The first thing I notice is that t is replaced with sin(t)

so we have 

\(x=sin^2t \qquad 0\le x\le 1\\ y=sin(t)+1 \qquad 0\le y \le 2\\ \mbox{So the domain and range are different.}\\ But\;\\ sin^{-1}(\pm\sqrt{x})=t \qquad and \qquad sin^{-1}(y-1)=t\\ \therefore\\ sin^{-1}(\pm\sqrt{x})=sin^{-1}(y-1)\\ \pm\sqrt{x}=(y-1)\\ x=(y-1)^2\\ x=(y-1)^2\\ \mbox{So this is exactly the same as the original}\\ \mbox{ relation except it has different a domain and range} \)

200000=100000e^.0475t    divide both sides by 100000

2 = e^[.0475t]     take the ln of both sides

ln 2  = ln e^[.0475t]      and by a log property we can write

ln 2  =  [ .0475t] * lne      the ln e  = 1......so we can drop this and we're left with

ln 2  = .0475t     divide both sides by .0475

ln 2 / .0475  = t   = about 14.59

to solve for t    first divide both sides by 100000

2=e^.0475t   take ln of both sides

ln 2 = .0475t

14.59=t

8/9 x 5/7 x 1/8

Here's another trick that sometimes works well......since we're just multiplying numerators/denominators, we can swap these around,  if we wish.....re-write this as :

8/8 x 5/7 x 1/9  =

1 x 5/7 x 1/9  =

5/7 x 1/9  =

5/63   [note that we didn't have to worry about "reducing" any fractions......]

5x + 7 = 3x - 5    subtract 3x from both sides

2x + 7  = -5          subtract 7 from both sides

2x  = -12           divide both sides by 2

x = -6

1 - 5 - 6 + 8 + 12  =

-4 - 6 + 8 + 12  =

-10 + 20  =

10

y = 2 ln (sec x)

dy/dx  = 2 [tan^2x]/ secx   = 2 [secx tanx] / secx   = 2 tan x

So at x = pi/4....we have

y '(pi/4)  = 2 tan (pi/4)   = 2 * 1   = 2

I don't know, EP......is this the thing that keeps asking you if you're a "robot?"

The first set of equations is

x = t^2   and   y = t + 1

Re-arranging the second one, we have y - 1  = t     and substituting this one into the first one, we have :

x = (y - 1)^2   which is a "sideways" parabola opening to the right with a vertex at (0, 1)

a) x = sin^2t    y = sint + 1

Rearranging the second one again, we have  sint = y - 1    and substituting this one into the first we have

x = (y - 1)^2 

However......this graph is a "restricted" version of the first one......from t = 0 to t = pi/2, the graph traces the same path from the vertex (0,1) to the point (1, 2).....then from t = pi/2 to t =pi.....it "reverses" course from (1,2) back to the vertex......from t = pi to t = 3pi/2, it traces the same path as the original graph from the vertex to the point (1, 0)...lastly, from t = 3pi/2 to 2pi......the graph again reverses course and ends back at the vertex.......this behavior is repeated infinitly

Here's the graph  :

b) x= t^4    y = t^2 + 1

Rearranging the second equation, we have y - 1 = t^2

Substituting this into the first equation we have the original equation, again, x = (y - 1)^2

However, this graph is just the [unbounded] upper half of the parabola since y is never less than 1

For  the interval where t ranges from (-infinity, 0), the graph traces back from the right to the point (0, 1).....when t ranges from (0, infinity), the gragh "reverses" course and traces back to the right

Here's the graph here traced from t = -2 to t = 2 :

Just multiply all of the numerators together and all of the denominators to gether    to get    40/504   = 5/63  (using the calculator!)

First you need to make sure you are working in consistent (the same) units

1.1 liter = 1100 cm^3 (cubic cm)

Volume of a cylinder   Vc = pi r^2 h   1100 = pi r^2 (27)

1100/(pi 27) = r^2    r=    3.601    Diameter = 2 x r  = 7.202 cm

6*8/12 +5 + 1 = 10  is another way

Before you can add or subtrac numbers with fractions involved you have to have a COMMON DENOMINATOR.

The denoms in this problem are   21   and 7          21  is an easy common denom for these tww.....right?

22  2/21  =   ((22 x 21) + 2 )  / 21 = 464/21          8 1/7  = ( (8x21) + (3x1) ) / 21  =  171/21

Now we can do the math:     464/21  - 171/21  =   (464-171)/21 = 293/21         you can simplify this to    13  20/21

How can you make 10 out of the numbers 1,5,6,8, and 12 using any or all operations?

Here is one way:

[12+8+1] - [6+5] =10

Hello, I'm a physics student at graduate level currently and that is a really nice and actually deep question. In my discipline math is viewed mostly as a language, as a universal (understood by every mathematically literate person) way of expressing events in nature. The thing is that math on its own is infallible, that means that its logic is perfect and the mathematical world does not always cather to reality. While you may have concepts in math as, let's say a mathematical point (0 dimensions), you cannot have such a thing in the physical reality as even the smallest of smallest things will actually have some real dimension (so far the "smallest" thing possible is thought to be the Planck length, though only theoretically).  Still, it is a very useful tool in I'd say every science. I have friends in biology and they use a ton of statistical analysis, some friends in chemistry that use a lot of complex mathematics as well. It's something that's just there, you will have to learn it if you want to do any physical science. Best believe it gets more interesting the more you do it!

Probabiltiy say 1 out of 10 tryout.....if 22 tryout then there is    10 x 22 in class = 220  

If t is in years and 315000 is the (in question) population in the present day, then t = 2020-2016 = 4 yrs from now

315000 (1.02)^4 = 340,966

n^4 - 100  =

(n^2 + 10) (n^2 - 10) =

(n^2 + 10) (n - √10) (n + √10)

Let x and y be the dimensions of the rectangle. Then from the given information \(\sqrt{x^2 + y^2} = 41\) and , so x + y = 49.

Squaring the equation \(\sqrt{x^2 + y^2} = 41\), we get \(x^2 + y^2 = 1681.\)

Squaring the equation x + y= 49, we get \(x^2 + 2xy + y^2 = 2401\)

Subtracting these equations, we get 2xy = 720, so the area of the rectangle is 360 degrees.

The missing side length of the first rectangle is 6 inches, because 30-(9+9)=12, and 12/2=6. (9+9+6+6=30) Now, 6 is equivalent to 2/3 of 9. Therefore, we can assume that the same baic principle applies to the second rectangle, based upon the information provided by the question. 

So, if the side length of one side of the rectangle is 6, we must assume that the length of the other side is 6* 2/3, or 4. 6+6+4+4=20, so we can assume that 20 is the length of the rectangle.

Dude that's so easy. You just subtract 148 from 6,666 to get 6518. LOL.

Theorems in daily life are used more commonly than you think. In fact you use at least one every day. Theorems are like temporary solutions to problems. Until something better comes up, that's how you should go about your observation or situaton. 

Math is the basis most -if not all- sciences! I wish I could explain but I'll let physics and chemstry do that...