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First one      find where x = 2     go up from there to intersect the red line....the 'y' value there is   f(2)

Second one     the blue graph is shifted UP   3 units from the red one   so    blue f(x)  = red f(x) + 3

Third one        the blue graph is shifted 2 units LEFT       blue f(x) = red f(x+2)

(10 + 5 + 5)(8 + 5 + 5) = 20 * 18 = 360 ft² 

360 - (10 * 8) = 360 - 80 = 280ft², which is your answer :)

Here is why (let O be the centre of the circle)

Consider minor arc AB

Angle ACB= half of angle AOB = 20 degrees 

{When standing on the same arc the angle at the centre will be twice the angle on the circumference.}

Consider minor arc CD

Angle CBD = half of angle COD = 84 degrees

Pink angle at E = 180-84-20 = 76 degrees.

I assume you know the Laplace transform of y`, y and a constant, and the inverse transform of 1/s and 1/(s+constant):

Here is an interactive drawing of it.  

How have you tried to tackle this guest?

For    \(0

\(\frac{(x-k)^2}{4}+\frac{y^2}{1}=1\qquad and \qquad \frac{(x+k)^2}{4}+\frac{y^2}{1}=1\)

What are the centres of each of these ellipses?

What can you say about these two ellipses?  What is the same and what is different?

So if you draw both of them on the same graph, what line will they have to cross on?

What is the length of the semi-major axis for each.

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What are the properties of a square?  Which might be the most helpful?

Sketch what I have talked about so far.  See what you can work out for yourself.

I would like you to interact with me. Tell me what you have done and ask questions.

The next step is yours.

This will help though.

x  = 50°

How many ways are there to give 9 posters to 2 friends.

Poster 1 can go to Aor B   2 ways

Poster 2  can go to A or B 2 ways

..

Poster 9 can go to A or B 2 ways

2^9 in total

We only need one  varaible  and only  one equation

T =  number  of tacos bought

T + 3  =  number of hotdogs bought

Cost of a taco * number bought  +  Cost of a hotdog * number bought  =  total cost

3.50 T   +  4 ( T + 3)  =  63

BC = DC   ....so...

CBD =  BDC   =  34° 

C (20, 3)   =   1140  ways

sin R  / r   = sin Q  /  q

sin R  /  940     =  sin 23  / 790

sin R  =  940 sin 23   / 790    

arcsin [   940 sin 23   / 790    ] =  R   = 27.7°

Second  possibility  for  R  =  180  - 27.7 =  152.3°

152.3  + 23  =  175.3   <  180......so....a second triangle  exists

∠ADC = 180 - 56

How many ways are there to divide a group of 6 friends among the basketball team, the soccer team, and the track team? (Each team could have anywhere from 0 to 6 of the friends on it. Assume the friends are distinguishable.)

It is very easy to double count in these questions so do not assume mine is fully correct.

If angles P and N are congruent, then OP and ON are congruent as well!!!

x = 1/2(360 -168 - 40)

x = 76 degrees

Wow! Thank you - I appreciate it "CPhill"!

Angle  J  =   180  - 160  - 14    =   6°

Using the  Law  of Sines

J  / sine  of angle J    =  L / sine  of angle  L

8.1  / sin 6° =  L  / sin 14°

L  =  8.1 sin 14°  / sin 6° =  18.7  inches

LOL half of 86 is 43 where I hail from.  :D 

Go  to  x  = 2

From there, go up  until you see  where  the graph  intersects a y value.....what is this y value???

That will be your answer, QM....let me know if you  get  stuck....

Find $f(2).$

sorry

IJ  = JK

So angles  JIK  and JKI are  equal  =  (180  -  94)  / 2  =  86 / 2  =   43° 

Find  ????

Welcome  !!!!

Note that  triangle  BYX  is similar  to triangle BCA

AC  =  BC  =   12 +  6   = 18

BA  =  sqrt  (18^2  + 18^2)  = sqrt (2 *18^2)  =   18 sqrt (2) 

Scale  factor  of   BYX  to  BCA  =  12/  (18 sqrt (2))  =  2/ (3 sqrt (2))  =  sqrt (2)  / 3

Ratio  of  area of BYX  to  BCA =   (scale  factor)^2  =   2/9

Just  as  QM  found  !!!!