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I have to add to get the answer?

OK.....I'll have to think a bit about that....!!!!!!

(☞ ͡ ͡° ͜ ʖ ͡ ͡°)☞ What do you think of that?

Since KLJ is a parallelogram, it can be split into 2 congruent triangles, as seen above. The angles remain the same, so we know angle KJL is also 25 degrees. Using the triangle angle sum theorem (all sides of a triangle add up to 180 degrees), we can determine the missing angle measure is 25 degrees. It would have been nice if the question told us JKLM was also a rhombus.

Cool! I didn't even see the fact the triangle was isosceles! I grant you one wish. (∩ ͡° ͜ʖ ͡°)⊃━☆

Angle ABC  = Angle ACB...so....AB = AC.....so 

4x + 4  = 6x - 14         add 14 to both sides, subtract 4x from both sides

18  = 2x    ⇒    x  = 9

So

BC  =   2x + 7   =    2*9 + 7  =  25

This is an isosceles triangle. That means the 2 sides touching the = angles have equal length. That means we know \(4x+4=6x-14\) .

To solve, we need to find x. I'll just quickly go through this: \(4x+4=6x-14\rightarrow4 = 2x - 14\rightarrow18=2x\rightarrow x=9\) .

Now that we have x by setting the 2 sides equal to each other, we know the length of the bottom side by substituting 9 in for x, giving us \(2x+7=2(9)+7=18+7\rightarrow x =24\) .

This triangle is isosceles   with  AB  = AC  ....so....the angles opposite these sides are also =

So  angle   (5x)°  = 40...so these two angles sum to 80°

So...this means that    2y + 20  must = 100

So

2y =  80

So....y  =  40

The sum of angles in a triangle adds up to 180 degrees. So we can subtract 60 to get 120 = 5x + 2y. Now we need to solve for y. Y is on the same side of the equation as x, so we need to move the x by subtracting 5x. Then we have 2y = 120 - 5x. Last, we need to divide by 2, as that is the coefficient of y. Now, we have y = 60 - 2.5x.

What is m∠C?

x + 2x + 3x  =  180

6x  =  180  divide both sides by 6

x = 30°

And   2x  =  60°  = angle C

If these are corresponding angles, the lines will be parallel.....and corresponding angles are equal.....so.....

3x + 5  =  65     subtract 5 from each side

3x  = 60   divide both sides by 3

x =  20

What is m∠C?

https://static.k12.com/nextgen_media/assets/8080707-NG_GMT_SemA_05_UT_01.png

For it to be parallel, 3x + 5 = 65

To solve the equation, x = 20

(12 * 7)  ( 4 + 1)  =

84   *   5   =

420    ????

The answer to a is pi * x

Use this formula to solve the second one, this will help you learn. 

3 * 2 * pi / 2 = 3pi

3/5 * 1/3 + 14/5 = 3

12 * 4 * 7 + 1

I think i wrote six...

Mmmmm....I'll try rosala ....!!!

Let's suppose that we have a partial series like this  where     a  = the first term   and d is the common difference between terms

a  + (a + d)   + (a + 2d)  + (a + 3d) + (a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)   + (a + 8d)

Let  n  be the number of even place terms

And    n + 1  the number of odd place terms

Then the sum of the even place terms is :

n *a   +   n^2* d     =  n (a + n*d)    

And the sum of  the odd place terms is : 

(n + 1) * a  + ( n+1) (n)* d   =  (n + 1) (a + n*d)     for  n ≥ 0  

So....the ratio of the sum of the odd place terms to the even place terms is :

(n + 1) (a + n*d)               ( n + 1)

_____________   =         _______

   n    (a + n*d)                       n

The goal is to get a common denominator and combine the fractions into one single fraction.

Notice that  (a - c)  is in the denominator of the first fraction, but not in the second.  So we need to multiply and divide the second fraction by  (a - c) . That is, multiply it by  (a - c)/(a - c)

\(\frac{1}{(2a-b)(a-c)}+\frac{1}{(b-c)(b-2a)} \\~\\ =\frac{1}{(2a-b)(a-c)}+\frac{1(a-c)}{(b-c)(b-2a)(a-c)}\)

Now we need to multiply the first fraction by  (b - c)/(b - c) .

\(=\frac{1(b-c)}{(b-c)(2a-b)(a-c)}+\frac{1(a-c)}{(b-c)(b-2a)(a-c)}\)

Now, notice that the term (2a - b)  is  -1 multiplied by  (b - 2a) .

So let's multiply the second fraction by  -1/-1 .

\(=\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{(-1)(a-c)}{(b-c)(-1)(b-2a)(a-c)} \\~\\ =\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{-a+c}{(b-c)(-b+2a)(a-c)} \\~\\ =\frac{b-c}{(b-c)(2a-b)(a-c)}+\frac{-a+c}{(b-c)(2a-b)(a-c)}\)        And distribute it.

Now the denominators are the same and we can add the fractions together.

\(=\frac{b-c+-a+c}{(b-c)(2a-b)(a-c)} \\~\\ =\frac{b-a}{(b-c)(2a-b)(a-c)}\)           -c  and  +c  cancel