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Try  (0,0)     

x^2+y^2-6x+4y+4=0       put into standard circle equation form

x^2 - 6x      +    y^2 + 4      = - 4       now, complete the square for x and y

(x-3)^2       +   ( y+2)^2     = - 4  + 9 + 4

(x-3)²         + ( y+2)²       = 9                     center is thus seen as   3, -2     and radius = sqrt 9 = 3

(2x -1 )(x  +3 ) = 0     shows x = 1/2  and  -3

This has nothing to do with quadratic functions asinus.

Hello Julia!

My proposition: answer c).

\(f(x)=sin^2(x)\\ g(x)=cos^2(x)\\ axis\ of\ reflection: y=0.5\)

  ?

I graphed it just becasue that is my idea of fun.

But really you are only interested in D  the other points have no consequence.

It is sensible to do a rough graph of D so that you can see where you might go with it.

D(4,1)

reflect it across the x-axis and you get   D'(4,-1)

Now you want to reflect that across the line y=x+1

the gradient of that line is one the line at right angles will have a gradient of -1  so we want the line that passes thorough (4,-1) with a gradient of -1

y=-x+b

-1=-4+b

b=3

y=-x+3

I also want the perpendicular distance from  (4,-1) to   y=x+1

x-y+1=0

\(d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}\\ d=\frac{|4+1+1|}{\sqrt{1+1}}\\ d=\frac{6}{\sqrt2}\\ d=3\sqrt2\)

So I want the point on the line y=-x+3 that is 3sqrt2  from (4,-1)

\((x-4)^2+(y+1)^2=72\quad   \cap \quad    y=-x+3\\ (x-4)^2+(-x+3+1)^2=72\\ (x-4)^2+(-x+4)^2=72\\ 2(x-4)^2=72\\ x-4=\pm\sqrt{36}\\ x=\pm6+4\\ x=-2 \;\;or \;\;10 \)

I obviously want the negative one

x=-2,  y=--2+3 = 5

so D" is (-2,5)

Think about it first, but then if you have questions, make sure you ask.    

I got (-2,5) is that correct?

25+1+1+1=28

is this a trick question? he decided to make enough Indian Pudding for everyone in his class to have half of a cup of pudding.

so maybe 25...............

nice! i didn't see it until now proyaop!

last point? Can you give a picture?

but if it looks like a right triangle, then since the top left point lines with the bottm left, there is exactly enough room for another right triangle in the rectangle

Properties of Rectangle say that the opposite sides are congruent, and all angles are 90 degrees. The diagonal is congruent to itself so by HL congruency theorem, we can say that the two triangles are congruent, because they are congruent they add up to the area of the rectangle, so that is why one of them has half the area of the rectangle. 

@ #3 guest 

i'm pretty sure you can solve that, pretty similar to the question asked

To simplify:

3sqrt5 + 2sqrt(5) + 6sqrt(5) = 11 sqrt5 so the answer is 11sqrt(5)

\(\sqrt{45}+2\sqrt5+\frac{\sqrt{360}}{\sqrt2}\)

\(3\sqrt5+2\sqrt5+\frac{\sqrt{360}}{\sqrt2}\)

\(3\sqrt{5}+2\sqrt{5}+6\sqrt{5}\)

\(11\sqrt{5}\)

By Vieta, the two roots add up to -b in the quadratic, as long as 'a' is 1. Thus, r + s = 6, so 2r + 2s = 12.

There are 3 roots each 2pi/3 = 4pi/6 apart

\(\sqrt[3]{8i}=-2i\)

-2i  has a anticlockwise angle of    \(\frac{3\pi}{2}=\frac{9\pi}{6}\\ \)

\(\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\ \frac{5\pi}{6}-\frac{4\pi}{6}=\frac{9\pi}{6}-\frac{4\pi}{6}=\frac{5\pi}{6}\\~\\ \text{so the 3 cubic roots are }\quad \\ 2e^{(\frac{\pi}{6})i},\quad 2e^{(\frac{5\pi}{6})i},\quad 2e^{(\frac{3\pi}{2})i}\\ \sqrt[3]{8i}=2(cos\frac{\pi}{6}+isin\frac{\pi}{6}),\;2(cos\frac{5\pi}{6}+isin\frac{5\pi}{6}),\;2(cos\frac{3\pi}{2}+isin\frac{3\pi}{2})\\ etc\)

checked:

Triangle ABC has the height AD and angle A = 90º. If CD = 4, and BD = 5, find this ratio [ACD]/[ABD].

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AD = x

√(4²+x²) * √(5²+x²) = 9x

AD = 2√5

[ACD] / [ABD] = 4/5

\(cos 30^\circ=-\frac{\sqrt3}{2}\)        (knowledge from a 30,60, 90 triangle)

2nd or 3rd quad

smallest is   180-30 = 150       next is    180+300=210

next is     150+360= 510   BINGO

Just as asinus said!  

You're welcome Julia.

You should get used to plugging you equations into Desmos AFTER you have solved them yourself.

This will increase your understanding of how the algebra and the graphical representations relate to each other.

You will enhance your learning enormously.

∠BAC = x              ∠DAC = 2x

2x + 10x - 22 = 180

Sylvia combines 8.65 ounces of strawberries with 8.09 ounces of blueberries to make fruit bowls. She pours the fruit equally into 4 bowls, and has 2.78 ounces of fruit left over. How manyounces of fruit are in each bowl?

Add the amount of strawberries to the amount of blueberries.

From this sum subtract the number of ounces of fruit that are left over.

To find the number of ounces of fruit in each bowl, divide the answer from step 2 by 3.

Strawberries =  8.15 / (8.15 + 6.26) * .73        then the rest would be blue

Hello >> Glenis Grace

https://www.youtube.com/watch?v=AdbZuj0dxHc