All Answers 358089 Answers

First and Second one graphs

Third One   Multiplying by 2 makes the wave TALLER....vertically stretched  by 2

                    Adding 3   shifts it UP 3 units

Second One

The high and low of the ferris wheel describe the DIAMETER

    High 55   Low 5       Diameter  50   radius 25 yds

Low point 5 (assumed to be ground level)  High point  55      50 yds  above the ground

Low point 9s    next low point  21 sec       takes 12 seconds for one rev

At t= 0   according to the table   ferris wheel height is 30 yds     Low point is at t=9s  5 yds

First one    See :  https://web2.0calc.com/questions/check-my-answer-please_1

3 C   take away 5 C

3-5 = -2 C

Thank you :) 

Q1)

Q2)

Q1)

There are two points, A and B.

The point A is (1,1) and point B is (5,6)

Point P is along the line segment of AB, and it makes a ratio of AP to BP to 1:2. 

What is point P?

Round to tenth decimal

\(\begin{array}{|rcll|} \hline \vec{P} &=& \vec{A}+ \lambda \left( \vec{B}-\vec{A} \right) \quad | \quad \lambda = \dfrac{1}{1+2} = \dfrac{1}{3} \\\\ \vec{P} &=& \vec{A}+ \dfrac{1}{3}\left( \vec{B}-\vec{A} \right) \quad | \quad \vec{A}=\dbinom{1}{1},\ \vec{B}=\dbinom{5}{6} \\\\ \vec{P} &=& \dbinom{1}{1}+ \dfrac{1}{3}\left( \dbinom{5}{6}-\dbinom{1}{1} \right) \\\\ \vec{P} &=& \dbinom{1}{1}+ \dfrac{1}{3} \dbinom{5-1}{6-1} \\\\ \vec{P} &=& \dbinom{1}{1}+ \dfrac{1}{3} \dbinom{4}{5} \\\\ \vec{P} &=& \dbinom{1+\frac{4}{3}}{1+\frac{5}{3}} \\\\ \vec{P} &=& \dbinom{ \frac{7}{3}}{ \frac{8}{3}} \\\\ \mathbf{\vec{P}} & \mathbf{=} & \mathbf{\dbinom{2.3}{2.7}} \\ \hline \end{array}\)

P = (2.3, 2.7)

Thank you!

Given that (2,3) is on the graph of y=f(x), find a point on the graph of y=f(2x+1)+3. 

Express your answer as an ordered pair (a,b) where a and b are real numbers.

\(\begin{array}{|rcll|} \hline y &=& f(x) \quad | \quad (x=2,\ y= 3 )\\ 3 &=& f(2) \\ \mathbf{f(2)} & \mathbf{=} & \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline y &=& f(2x+1)+3 \\\\ 2x+1 &=& 2 \\ 2x &=& 1 \\ \mathbf{x} & \mathbf{=} & \mathbf{\dfrac{1}{2}} \\\\ y &=& f(2) + 3 \quad | \quad \mathbf{f(2)=3} \\ y &=& 3 + 3 \\ \mathbf{y} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}\)

\((a,b) = \left( \dfrac{1}{2},\ 6 \right) \)

10.5?

That is cute!  I will try and remember that one :)

You can plug your answer into both equations and check if they are true.

i think the answer might be -6  and -4

1. Multiply second equation by 2

\(\frac{5}{3}\)x-2y=-2

-2x+2y=4

2. Use elimination to cancel the y's

-2x+\(\frac{5}{3}\)x=2

3. Now its your turn

Solve the equation for x, plug it back in one of the equations, then you get your answer.

(I want you to do it yourself so you learn)

OK mclovin....these are all the same...just different numbers

equation 2 says that   y = 2+x       Put that in equation 1     solve for x    then use this value of x in one of the equations to find y

try it  

Take the SECOND equatiion and re-arrange:

y = -3 - x     Now substitute this into the first equation to solve for  x     this is jsut like your last question,,,,,just different numbers...

                  try it ....post again if you need more help

These are equations for lines......where they cross (intersect) is the solution to the system......straight lines only intersect at one point....one solution.

ohhhhh i want to facepalm myself now. thank you!

"five"

remove 2, the "f" and the "e"

"iv" = 4 in Roman numerals

Sorry i didn't see your reply

oooh nvm ok thank you