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The product of 1 multiplied by 2 is simply 2. In mathematical terms, when you multiply a number by 1, the result is always the original number itself. Therefore, 1 times 2 equals 2.

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2w ago
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12y ago

A monomial expression.

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Q: What is 1 x2?
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What is x2 plus x2 plus x2?

x2 + x2 + x2 = (1 + 1 + 1)x2 = 3x2


What is the anti derivative of x divided by x squared minus 1?

The antiderivative of x/(x2-1) is ln(x2-1)/2. Proof: (ln(x2-1)/2)' = (1/(x2-1))*(x2-1)'/2=1/(x2-1)*(2x/2)=x/(x2-1).


Sum of series 1 plus x2 plus x4.....plus xn?

(xn+2-1)/(x2-1)ExplanationLet Y=1+x2+x4+...+xn. Now notice that:Y=1+x2+x4+...+xn=x2(1+x2+x4+...+xn-2)+1Y+xn+2=x2(1+x2+x4+...+xn-2+xn)+1Y+xn+2=x2*Y+1Y+xn+2-x2*Y=1Y-x2*Y=1-xn+2Y(1-x2)=1-xn+2Y=(1-xn+2)/(1-x2)=(xn+2-1)/(x2-1)


How do you factorise x2-1?

1(x2-1)


X6 plus 3x4-x2-3 equals 0?

x6 + 3x4 - x2 - 3 = 0(x6 + 3x4) - (x2 + 3) = 0x4(x2 + 3) - (x2 + 3) = 0(x2 + 3)(x4 - 1) = 0(x2 + 3)[(x2)2 - 12] = 0(x2 + 3)(x2 + 1)(x2 - 1) = 0(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0x2 + 3 = 0x2 = -3x = ±√-3 = ±i√3 ≈ ±1.7ix2 + 1 = 0x2 = -1x = ±√-1 = ±i√1 ≈ ±ix + 1 = 0x = -1x - 1 = 0x = 1The solutions are x = ±1, ±i, ±1.7i.


Factor of 1-x4?

1 - x4 = (1 - x2)(1 + x2) = (1 - x)(1 + x)(1 + x2) (difference of squares)


How do you completely factorise x to the fourth power minus x to the third power minus x minus 1?

x4 - x3 - x - 1 rewriting: = x4 - 1 - x3 - x factorising pair of terms: =(x2 + 1)*(x2 - 1) - x*(x2 + 1) = (x2 + 1)*(x2 - 1 - x) or (x2 + 1)*(x2 - x - 1) which cannot be factorised further.


What is the integral of rootsquare 1 plus x2 over x without using partial fraction?

Int sqrt(1+x2)/x = sqrt(1+x2) + LN [(sqrt(1+x2) - x -1) / (sqrt(1+x2) - x +1)]


How do you do recursive pattern rule?

a recursive pattern is when you always use the next term in the pattern... for example 4,(x2+1) 9,(x2+1) 19,(x2+1) 39,(x2+1) 79,(x2+1) 159


Expand x2 - 12x3?

x2 - 12x3 = x2(1 - 12x)


Convert x2 plus y2 equals 2x?

What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2


How do you swap two numbers without third variables?

If the variables are x1 & x2 the solution is : 1) x1=x1+x2; 2) x2=x1-x2; 3) x1=x1-x2; EX: x1=1 , x2=6; 1) x1= 1+6 = 7 2) x2= 7-6 =1 3 x1=7-1 =6 ============================================