quinn

in the isosceles triangle ABC, segment AD is the height of triangle from point A,and segment BE is the height of the triangle from point Learn from point B.

because triangle ABC is a isosceles triangle ,so AB=AC ,and segment AD is the height of the triangle 

so angle ADB= Angle ADC=90 degrees

In the right triangle  ADB and right triangle ADC 

AB=AC,AD=AD, so right triangle ADB is congruent to right triangle ADC ( hypotenuse-leg)

so angle BAD= Angle CAD

If angle BAD=x degrees ,then angle BAC= 2x degrees

AD=AB*cosx  DB=AB*sinx   BE=AB*sin2x

the area of triangle = BD *AD*2/2=BD*AD=BE*AC/2=BE*AB/2

BD*AD=BE*AB/2

AB*cosx *AB*sinx=AB*sin2x*AB/2 (cancel AB*AB from both side and both side times 2)

2*cosx*sinx=sin2x

this is my method,Can someone prove it algebraicallly?

okay! Thank you,guys.

-2*4/3=-8/3

0.1414=14.14%

okay the original question is In a trilsupose a lawyerwants 2 men and 5 womento make up a special panel.if the 7 panel  menbers are selected at random fro a pool of 12 peoples,which istheprobality that the  lawyer will not get the desired panel?(sorry for i didnt write the question completelyand i got the anwser）

ok i got it know.Thank you for your help.

why 0!=1? can n be a negative or fraction in n! ?why or why not?

i missunderstood the word "diameter".I got it now.

Thank you,geno3141.

This is a kind of ACt pratice problem in Mississippi ACT Exam Prep workbook. Accctually, the triangle ABC has a height of 1.5 units.But i think it can not be,becasue the height of the triangle ABC should be shorter than the radius.And the area of the triangle ABC is 2.25 units square.So the base of the triangle ABC is 3 units.I am wondering how they get 3.I put the problem into the forum because i think they did wrong on this problem. And  i want to find ot the anwser togerther.

the slope intercept form of linear function is y=mx+b

so in number 1 ,slope =1,so m=1 ,y=x+b,Then plug in (0,1）(x=0,y=1），1=0*1+b,b=1,so the equation of number 1 is y=x+1

2.   y=-3x-2/5

3    y=3x/4+2

4    y=-1x/5+1

this is my method:

2a^2+2b^2+2c^2=2ac+2bc+2ac    (both side times two)

a^2+2ab+b^2+b^2+2bc+c^2+a^2+2ac+c^2=0

(a-b)^2+(b-c）^2+(a-c)^2=0

because the square of all real number always more than 0

so (a-b)^2=0 , (b-c)^2=0 ,(a-c)^2=0

so a=b=c

so this is a Equiatreral Triangle

Good job guys!

you guys can do the second one in this way

(x+1)(x+7)(x+3)(x+5)+15

(x^2+8x+7)(X^2+8x+15)+15

(x^2+8x)^2+22(x^2+8x)+120

(x^2+8x+10)(x^2+8x+12)

(x^2+8x+10)(x+2)(x+6)

Good job Melody and CPchill