PLSS HELP ME CUS YES
2 answers:
You might be interested in
Sum of 2 perfect cubes
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
a³+b³=(a+b)(x²-xy+y²)
so
x³+4³=(x+4)(x²-4x+16)
set each to zero
x+4=0
x=-4
the other one can't be solveed using conventional means
use quadratic formula
for
ax^2+bx+c=0
x=
for x²-4x+16=0
x=
x=
x=
x=
x=
x=
the roots are
x=-4 and 2+2i√3 and 2-2i√3
Answer:
Step-by-step explanation:
( Being vertically opposite angles)
Vertically opposite angles are always equal.
Move variable to L.H.S and change it's sign
Similarly, Move constant to R.H.S and change it's sign
⇒
Collect like terms
⇒
Subtract 18 from 50
⇒
Divide both sides of the equation by 2
⇒
Calculate
⇒
The value of x is 16
Now, let's find value of m<AFD :
( sum of angle in straight line )
plug the value of x
⇒
Multiply the numbers
⇒
Add the numbers
⇒
Move constant to R.H.S and change it's sign
⇒
Subtract 98 from 180
⇒
Value of m<AFD = 82
Hope I helped!
Best regards!!