<u>Answer </u><u>:</u>
In the given quadrilateral ABCD ,
- Angle BCA = 18°
- Angle ACD = 62°
Angle BCA = Angle CAD ( alternate interior angle )
Now in triangle CAD ,
We have two angles so by using angle sum property we can find the required third one ,
- Angle CAD + Angle ACD + Angle ADC = 180°
- 18 + 62 + Angle ADC = 180
- 80 + Angle ADC = 180
- Angle ADC = 180 - 80
- Angle ADC = 100
Answer:
C
Step-by-step explanation:
It can be any combination from 2/3/5/8
Answer:
-3.5,-0.2,0.6,5 ... is the answer
Just like at any other time, to add/subtract fractions you need a common denominator.
14)
1/(x^2+2x)+(x-1)/x=1 so we need a common denominator of x(x^2+2x)
[1(x)]/(x(x^2+2x))+[(x-1)(x^2+2x)]/(x(x^2+2x))=[1(x(x^2+2x))]/(x(x^2+2x))
now if you multiply both sides of the equation by x(x^2+2x) you are left with:
x+(x-1)(x^2+2x)=x(x^2+2x)
x+x^3+2x^2-x^2-2x=x^3+2x^2
x^3+x^2-x=x^3-2x^2
x^2-x=-2x^2
3x^2-x=0
x(3x-1)=0, x=0 is an extraneous solution as division by zero is undefined. So the only real solution is:
x=1/3
...
16)
(r+5)/(r^2-2r)-1=1/(r^2-2r) the common denominator we need r^2-2r so
[r+5-1(r^2-2r)]/(r^2-2r)=1/(r^2-2r), multiplying both sides by r^2-2r yields:
r+5-r^2+2r=1
-r^2+3r+5=1
-r^2+3r+4=0
r^2-3r-4=0
(r-4)(r+1)=0, r^2-2r cannot equal zero, r(r-2)=0, r cannot equal 0 or 2...
r=-1 or 4
