If X=5..... x4+6-8=?

Answer 1

18

Answer 2

x=5
x4+6-8=?
(5)4+6-8=?
20+6-8=?
26-8=?
=18

always remember PEMDAS (Parenthesis Exponents Multiplications Divisions Addiction and Subtraction)

Answer 3

Use BODMAS (Brackets of Division, Multiplication, Addition, Subtraction) (Some call it BOMDAS)

X = 5
X=5 is the same as saying (5)4+6-8 = ?

5(4) = 20 (5 multiplied by 4)
20+6 = 26
26 - 8 = ?

Your answer is 18.

Answer 4

18

Answer 5

18

5(4) is 20
20+6 = 26
26-8 = 18

Answer 6

eh..

x=5 is the same as saying (5)4+6-8 = ?

5(4) = 20
20+6 = 26
26 - 8 = ?

(gotta leave something for you to do)

Answer 7

i haven't bothered to learn....is the unit step function H(x) considered "problem-loose?" i'm particularly particular it has a ability sequence strengthen that converges properly adequate. i will use the convention that H(0) = 0. yet besides, forgive the complexity... F(x,y,z) = x*{H(x-y) + H(x-z) - H(y-z) - H(z-y) } + y*{H(y-z) + H(y-x) - H(x-z) - H(z-x) } + z*{H(z-y) + H(z-x) - H(y-x) - H(x-y) } interior the (possibly) trivial case that x = y = z , F(x,y,z) ought to return 0, suited because NONE are maximum excellent. EDIT, ok i theory-approximately yet in a diverse way this could behave badly, will behave badly....forgive much greater complexity. F(x,y,z) = x*H{ H(x-y) + H(x-z) - H(y-z) - H(z-y) } + y*H{ H(y-z) + H(y-x) - H(x-z) - H(x-z) } + z*H{ H(z-y) + H(z-x) - H(y-x) - H(x-y) } with a bit of luck that association of H()'s will return a million provided that the respective variable is the terrific, in no way -a million

Answer 8

Apply DMAS rule to solve equations like that

1st D divide
2nd M multiply
3rd A add
4th S subtract

Answer 9

18............we just went over this in math class