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First   multiply   2 x     *  (5x-y+3)   =   10x^2 - 2xy+6x

  then multiply      y     *  (5x-y+3)   =                5xy        - y^2  + 3y  

then add the REDs together......

Find the lengths of kite's diagonals using the Pythagorean theorem. Example:   AC = sqrt( 9² + 12² )

When you have the lengths of both diagonals, multiply them and divide their product by two.     

I won't solve this simple geometric problem for you, but I'll give you a hint:

In this case, the height is also a median. Find the height, and then find SX using the Median theorem.

Important::::The Median Theorem states that the medians of a triangle intersect at a point called the centroid that is two-thirds of the distance from the vertices to the midpoint of the opposite sides.

This is given      -2 (x-3-sqrt5)(x-3+sqrt5)      multiply it out and reduce by collecting like terms

             and then you will see what  A  and  B    and C    are

   the 'x' of the vertex  is then   - B / 2A

       use this value of 'x' in the equation to compute the 'y' value of the vertex.....

The shortest leg is th eone between   1,12   and  7,10

  Use distance formula between to points to compute

(x1 -x2)^2  + (y1-y2)^2 = d^2

Cross multiply the L

sin^2 + 1 +2cos + cos^2 = 4 sin (1+cos)           edited

2 + 2cos = 4 sin(1+cos)

2 = 4sin

1/2 = sin    = 30, 150 degrees

The length of the dashed line is  1 + sqrt(3)  

Thank you for the diagram!!!

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1 oz is already missing   + 38 oz =   39 ounces  empty

39/50 x 100% = ........% empty

perimeter will be multiplied by 3.5       28 x 3.5 = ........

Area will be enalarged by   3.5 x 3.5      3.5² x (6*8) = .....

For the community: Please don't submit solutions to this problem. This is a homework problem for an online course that does not allow students to search for answers to specific problems outside of the school.

For the original poster: We realize that homework can be challenging. If you wish to receive help from the staff or other students, we encourage you to use the resources that the online classes provide. Please don't ask or search online for homework help. We understand that it's common in today's information age to look for resources online, and in some contexts, that's a great thing! However, it's against our Honor Code. You can ask for help on the message boards, and you can learn from the official solution after you submit your answer.

it says they are wrong :/ I'm still trying to figure it out

For the community: Please don't submit solutions to this problem. This is a homework problem for an online course that does not allow students to search for answers to specific problems outside of the school.

For the original poster: We realize that homework can be challenging. If you wish to receive help from the staff or other students, we encourage you to use the resources that the online classes provide. Please don't ask or search online for homework help. We understand that it's common in today's information age to look for resources online, and in some contexts, that's a great thing! However, it's against our Honor Code. You can ask for help on the message boards, and you can learn from the official solution after you submit your answer.

For the community: Please don't submit solutions to this problem. This is a homework problem for an online course that does not allow students to search for answers to specific problems outside of the school.

For the original poster: We realize that homework can be challenging. If you wish to receive help from the staff or other students, we encourage you to use the resources that the online classes provide. Please don't ask or search online for homework help. We understand that it's common in today's information age to look for resources online, and in some contexts, that's a great thing! However, it's against our Honor Code. You can ask for help on the message boards, and you can learn from the official solution after you submit your answer.

Hey, it's me again.  I was looking at the picture and an idea fell out of the sky. 

I don't know how to draw on this site, so I'll try to explain it in words. 

You have the large square which is the outside one. 

You have the small square which is the inside one. 

Draw a line from point A to point C and 

draw a line from point D to point B. 

Now you have four tiny squares, each of which is cut into two equal triangles by a diagonal line. 

The small square contains four of those triangles and   

the large square contains eight of those triangles. 

Thus, the large square contains twice the area of the small square. 

The area of the small square is 60, so twice that is the area of the large square, i.e., 120 

Neat, huh?  Even if I do say so myself.  TTFN. 

_.

oh okay thank you! I am going to look at the answers to understand it. thank you! I shouldn't have reposted.

This question has been posted here many times. See 3 different answers, all with the same probability, here:

https://web2.0calc.com/questions/let-triangle-abc-be-equilateral-where-the-side-length#r2
https://web2.0calc.com/questions/help-probability_2
https://web2.0calc.com/questions/help_1095

Distance to tower from building    tan 45 = d/7      solve for d = 7 m

Distance to top of tower from 7 m up     tan 60 = h/7    h = 12.12 m

    add h + the 7 m =   height = 19.12 m

There are 166 pairs that begin like this (a, b): (497, 2) , (494, 4) , (491, 6) , (488, 8) , (485, 10) , (482, 12) ....etc.

Notice that as "a" DECREASES by 3, "b" INCREASES by 2. This pattern holds for 166 pairs, the last one being:

a = 2  and  b =332.

Great job guest with two and three, I give a shot at number one. :)

So I am assuming that in LaTeX, \(\frac{s}{2}+5t=3, 3t-6s=9\) this is the system,

We can do substituiton or elimination, but I will do substitution.

Isolate s for The first equation to get \(s=6-10t\),

Substitute \(6-10t\) for s into the second equation to get \(\begin{bmatrix}3t-6\left(6-10t\right)=9\end{bmatrix}\),

Simplify into \(63t-36=9\),

And from here, I think that Caffeine should be able to do it him/herself. :)

x = 3 , -3               ( you would THINK     x = 2   would work, but   2 >= -4   so f(2) = 2²-4=0 )

so f(x)  has to equal 3  or -3

x^2-4 = 3       then x  = +-sqrt7

x^2-4 = -3      then x  = +- 1           4 answers I believe........                  edited

What do you want to do with this expression?