To solve the exercirses which are shown in the figure attached, you must follow the proccedure below:
7) (x)=x³-6x²+8x
(x)=x(x³-6x²+8)
(x)=(x-4)(x-2)x
The lenght is: x
The height is= (x-4)
8) √(2x+8)-6=4
1. You need to clear the variable "x". Then:
√(2x+8)=4+6
√(2x+8)=10
(√2x+8)²=10²
2x+8=100
2x=100-8
x=92/2
x=46
9) l4x+3l=9+2x
1. To solve the left member, you must evaluate two cases: it could be positive,or negative. Then:
2. Negative:
l4x+3l=9+2x
-4x-3=9+2x
-4x-2x=9+3
-6x=12
x=12/-6
x=-2
3. Positive:
l4x+3l=9+2x
4x+3=9+2x
4x-2x=9-3
2x=6
x=6/2
x=3
You labeled the triangle wrong sides 'a' and 'b' are supposed to be the sides that make the right angle. the other side is called the hypotenuse which is the longest side which you should have labeled 'c'
so Pythagorean theorem says
a^2+b^2=c^2
so
(2x+1)^2+(11x+5)^2=(12x+1)^2
distribute
(4x^2+4x+1)+(121x^2+110x+25)=(144x^2+24x+1)
add like terms
125x^2+114x+26=144x^2+24x+1
subtract 125x^2 from both sides
114x+26=19x^2+24x+1
subtract 114x from both sides
26=19x^2-90x+1
subtract 26 from both sides
0=19x^2-90-25
factor
(x-5)(19x+5)=0
therefor x-5=0 and/or 19x+5=0
so
x-5=0 add 5 to both sides
x=5
19x+5=0
subtract 5 from both sides
19x=-5
divide both sides by 19
x=-5/19
since side legnths can't be negative, we can cross this solution out
so x=5
subtitute
1+2x
1+2(5)
1+10=11
side a=11
11x+5
11(5)+5
55+5=60
side b=60
12x+1
12(5)+1
60+1=60
side c=61
add them all up
side a+b+c=11+60+61=132=total legnth
1 quart I think tbh idk but i think .
