The 3rd one because of the laws and newton’s ways
8/x+9-(2/x-3)=0
Step by step solution :
Step 1 :
2
Simplify —
x
Equation at the end of step 1 :
8 2
(— + 9) - (— - 3) = 0
x x
Step 2 :
Rewriting the whole as an Equivalent Fraction :
2.1 Subtracting a whole from a fraction
Rewrite the whole as a fraction using x as the denominator :
3 3 • x
3 = — = —————
1 x
Equivalent fraction : The fraction thus generated looks different but has the same value as the whole
Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator
Adding fractions that have a common denominator :
2.2 Adding up the two equivalent fractions
Add the two equivalent fractions which now have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
2 - (3 • x) 2 - 3x
——————————— = ——————
x x
Equation at the end of step 2 :
8 (2 - 3x)
(— + 9) - ———————— = 0
x x
Step 3 :
8
Simplify —
x
Equation at the end of step 3 :
8 (2 - 3x)
(— + 9) - ———————— = 0
x x
Step 4 :
Rewriting the whole as an Equivalent Fraction :
4.1 Adding a whole to a fraction
Rewrite the whole as a fraction using x as the denominator :
9 9 • x
9 = — = —————
1 x
Adding fractions that have a common denominator :
4.2 Adding up the two equivalent fractions
8 + 9 • x 9x + 8
————————— = ——————
x x
Equation at the end of step 4 :
(9x + 8) (2 - 3x)
———————— - ———————— = 0
x x
Step 5 :
Adding fractions which have a common denominator :
5.1 Adding fractions which have a common denominator
Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:
(9x+8) - ((2-3x)) 12x + 6
————————————————— = ———————
x x
Step 6 :
Pulling out like terms :
6.1 Pull out like factors :
12x + 6 = 6 • (2x + 1)
Equation at the end of step 6 :
6 • (2x + 1)
———————————— = 0
x
Step 7 :
When a fraction equals zero :
7.1 When a fraction equals zero ...
Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.
Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.
Here's how:
6•(2x+1)
———————— • x = 0 • x
x
Now, on the left hand side, the x cancels out the denominator, while, on the right hand side, zero times anything is still zero.
The equation now takes the shape :
6 • (2x+1) = 0
Equations which are never true :
7.2 Solve : 6 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Answer:
here look for it there is a sol
Step-by-step explanation:
https://www.careerlauncher.com/cbse-ncert/class-9/Math/CBSE-SurfaceAreasandVolumes-NCERTSolutions.html
Evaluating the expression 11C3 is very simple to do.
If you are assuming that C is the notation for the combination.
11!
11C3= ____
3!(11-3)!
∠BDC=30, so ∠CBD + ∠BCD=180-30=150
ΔDBC is isosceles, so ∠CBD=∠BCD=half of 150=75
∠ABC=∠ABD-∠CBD=155-75=80
ΔABC is isosceles, so ∠ABC=∠ACB=80
∠BAC=180-80-80=20
