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I don't think that's correct

This is easy....the centroid is given by :

[ (  add the x coordinates of the vertices ) / 3, ( add the y coordinates of the vertices ) / 3 ]

So we have

[   ( -4 + 2 + 2 ) / 3  ,  ( -1 + 2 + -3 ) / 3  ]

[  0 / 3 ,  - 2/3  ]  

( 0 , - 2/3 )

Since angle ABC  = angle BCA....the sides opposite these are also equal...so

5x + 11  =  8x  - 4       add 4 to both sides, subtract 5x from both sides

15  =  3x        divide both sides by 3

5  = x

So....BC  = 4(5) - 2  = 20 - 2   = 18 

And yes, I was reffering to the tube link

Let C  =  (-3, -2)

Let A = ( 5, -2)

Let A' =  ( -5, - 2)

So we have that

< A'  > =   < A  - C > *(Scale Factor) + < C >

< -5, -2 >  =  <  8, 0 > *(Scale Factor) + < -3, -2 >

< -2, 0 >  =  < 8, 0 > * (Scale Factor )

So

-2  =  8 * Scale Factor     ....... divide both side by 8

- (1/4)  =  Scale Factor

Yes,  That was very good, 

I did understand the explination,

Sorry.....If you want a numerical answer... note that all the angles = 60°....so.....

25x - 15  =  60     add 15 to both sides

25x  =  75      divide both sides by 25

x  = 3

AB = AC.....so  angle ABC  = ACB...so each angle = 25°

So  the remaining angle = 180 - 25 - 25  =  130°

So

3y + 70  =  130     subtract 70 from both sides

3y  = 60     divide both sides by 3

y  = 20

So, it would be 27.5? 

We love your funny pics, Rosala.....!!!

This triangle is equilateral....so....all sides and angles are equal

So

2y  = 25x - 15    add 15 to o both sides

2y + 15   =  25x     divide both sides by 25

[ 2y + 15 ] / 25    = x

The circumcenter is where the perpendicular bisectors of the triangle meet

A(−3, 3) , B(−1, 3) , and C(−1, −1)

Find the midpoint  of  AC  =  (-2, 1)

And the slope between thes two points is  -2

And the slope of a perpendicular line is  1/2

So....the equation of the line that goes through (-2, 1)  with a slope of 1/2 is

y = (1/2)(x + 2) + 1

y = (1/2)x + 2      (1)

And the midpoint of  AB  is  ( -2, 3)

And the slope between these points is  0

So the equation of a perpendicular line to this one that goes through  (-2,3) is x = -2   (2)

So...putting  (2) into (1)  we have

y = (1/2)(-2) + 2

y = -1 + 2

y = 1

So....the circumcenter is  ( -2, 1)

Here's the graph :  

3x+2*[x-8]-12

3x + 2x - 16 - 12  =

5x - 28

Answer to Question 3:
The triangles are similar if the corresponding angles are EQUAL;
So, angle B = angle Q, angle A=angle P, angle C=angle R.
And the corresponding sides of the two triangle must be PROPORTIONATE, namely:
32/24 =28/[3x + 6]   cross multiply
24 x 28 = 32 x [3x + 6]
672 = 96x + 192
96x =672 - 192
96x = 480  divide both sides by 96
x = 480 / 96
x = 5  sub 5 for x in 3x + 6
(3 x 5) + 6 = 21 Length of the corresponding side with x.

hi hope your doing good  like me have a nise day

Solve for x:
x^4 + 2 x^2 - 3 = 0

Substitute y = x^2:
y^2 + 2 y - 3 = 0

The left hand side factors into a product with two terms:
(y - 1) (y + 3) = 0

Split into two equations:
y - 1 = 0 or y + 3 = 0

Add 1 to both sides:
y = 1 or y + 3 = 0

Substitute back for y = x^2:
x^2 = 1 or y + 3 = 0

Take the square root of both sides:
x = 1 or x = -1 or y + 3 = 0

Subtract 3 from both sides:
x = 1 or x = -1 or y = -3

Substitute back for y = x^2:
x = 1 or x = -1 or x^2 = -3

Take the square root of both sides:
x = 1    or    x = -1    or    x = i sqrt(3)    or    x = -i sqrt(3)

answer to Question 1: Look up the answer here:

http://web2.0calc.com/questions/proofs-it-has-a-number-6-and-7-i-just-couldn-t-fit

Answer to Question 2:

1. Open the compasses to a distance greater than half the distance between F and G.

2. Put the compass point on F and draw a semicircle that crosses line FG.

3. Repeat 2. with the compass point on G.

4. The two semicircles will intersect at two points, so use the straight edge to draw a line between them.  This line will be perpendicular to line FG.

Alan Nov 19, 2017

Lol this is funny...

but off topic

Hello!

I like your username.

There are few things to look for to determine whether the end behavior is "down and down, up and down, up and up."

1. Look at the Degree of the Polynomial Function

If the degree is odd, then the function will behave in an "up-down" behavior, if that is how you would like to think about it.

If the degree is even, then you will have to check one more thing.

2. If the Degree is Odd, then Look at the Leading Coefficient

The leading coefficient is the coefficient of the highest-degree term in the polynomial function. 

If the leading coefficient is positive, the graph will have an "up-up" behavior.

If the leading coefficiennt is negative, then the corresponding graph will have a "down-down" behavior.

Hope this helps!

I don't understandwhat I'm doing wrong. What does the (a) stand for in the function

\( f(a) = 3a – a^2\)

a is a variable and can be equal to any number or any mathematical expression.

It is the same as 

\(f(x) = 3x – x^2\\ f(x) = – x^2+3x\\ f(x) =-x(x-3) \)

This would graph as a concave down parabola with roots of 0 and 3

The points $(0,4)$ and $(1,3)$ lie on a circle whose center is on the $x$-axis.
What is the radius of the circle?

Let \(P_1 = (x_1,y_1) = (0,4)\)
Let \(P_2 = (x_2,y_2) = (1,3)\)
Let Center of the circle \(C = (x_c,y_c) = (\ ?,0)\)
Let radius of the circle = r

\(\begin{array}{|lrcll|} \hline & (x_c-x_1)^2 + (y_c-y_1)^2 &=& r^2 \quad & | \quad y_c = 0 \\ & (x_c-x_1)^2 + (0-y_1)^2 &=& r^2 \\ (1) & x_c^2-2x_cx_1+x_1^2 + y_1^2 &=& r^2 \\ \hline & (x_c-x_2)^2 + (y_c-y_2)^2 &=& r^2 \quad & | \quad y_c = 0 \\ & (x_c-x_2)^2 + (0-y_2)^2 &=& r^2 \\ (2) & x_c^2-2x_cx_2+x_2^2 + y_2^2 &=& r^2 \\ \hline \end{array} \)

\(\small{ \begin{array}{|lrcll|} \hline (1)-(2): & x_c^2-2x_cx_1+x_1^2 + y_1^2-(x_c^2-2x_cx_2+x_2^2 + y_2^2) &=& r^2-r^2 \\ & x_c^2-2x_cx_1+x_1^2 + y_1^2-x_c^2+2x_cx_2-(x_2^2 + y_2^2) &=& 0 \\ & -2x_cx_1+x_1^2 + y_1^2 +2x_cx_2-(x_2^2 + y_2^2) &=& 0 \\ & 2x_c(x_2-x_1)+x_1^2 + y_1^2 -(x_2^2 + y_2^2) &=& 0 \\ & 2x_c(x_2-x_1) &=& (x_2^2 + y_2^2)-(x_1^2 + y_1^2) \\ & \mathbf{ x_c } & \mathbf{=} & \mathbf{\dfrac{(x_2^2 + y_2^2)-(x_1^2 + y_1^2)} {2(x_2-x_1)}} \\ & && | \quad x_1 = 0 \quad y_1 = 4 \quad x_2 = 1 \quad y_2 = 3 \\ & x_c & = & \dfrac{(1^2 + 3^2)-(0^2 + 4^2)} {2(1-0)} \\ & x_c & = & \dfrac{10-16 } {2 } \\ & \mathbf{x_c} &\mathbf{ =} &\mathbf{ -3} \\\\ \hline (1) & x_c^2-2x_cx_1+x_1^2 + y_1^2 &=& r^2 \\ & (-3)^2-2(-3)(0)+0^2 + 4^2 &=& r^2 \\ & 9+16 &=& r^2 \\ & 25 &=& r^2 \\ & \mathbf{5} &\mathbf{ =}& \mathbf{r} \\ \hline \end{array} }\)

Hi ProMagna,

Well we cannot give you the exact answer if you do not give us the exact question...

OR were you referring to the you tube link?

Did you understand my explanation?