What is the answer to 2E2x2E-4

Question

It's on properties of integer exponents

Answer 1

Multiply the numbers but add exponents: 2*2E(2-4)=4E-2=0.04

Answer 2

64.
2^2=2*2=4
2^-4=-2*-2*-2*-2=16
16*4=64

Answer 3

Given: 2E2 x 2E-4   These two figures are written in E-notation.

2E2 indicates 2x10^2 in exponential/power-of-ten notation, and 2E-4 does 2x10^-4,so
2E2 x 2E-4=(2x10^2)(2x10^-4) ··· Eq.1

First, multiply two coefficients in front of exponentials, 2 times 2, and then multiply exponential components, 10^2 times 10^-4, and finally multiply the two products obtained.

Use the rules for exponents: a^m x a^n=a^(m+n) where a,m and n are any real
numbers, and a^-m=1/a^m (a is not 0), so Eq.1 can be rewritten:

2E2 x 2E-4=(2 x 2)(10^2 x 10^-4)=4x10^(2-4)=4x10^-2=4/10^2=4/100=0.04

The answer is: 2E2 x 2E-4=4E-2 (=4x10^-2=0.04)

The solutions shown above are all for explanations.   You can solve this question just multiplying coefficients, and then exponentials separately, e.g.

2E3 x 6E-5=(2x6)E(3-5)=12E-2=1.2E-1 (=1.2x10^-1=0.12) 

Answer 4

Given: 2E2 x 2E-4   these two figures are written in E-notation.
2E2 indicates 2x10^2 in exponential/power-of-ten notation, and 2E-4 does 2x10^-4, so

2E2 x 2E-4=(2x10^2)(2x10^-4) ··· Eq.1*

First, multiply two coefficients in front of exponentials, 2 times 2, and then multiply the exponential components, 10^2 times 10^-4, and finally multiply the two products obtained.

Use the rules for exponents: a^m x a^n=a^(m+n) where a,m and n are any real numbers, and a^-m=1/a^m (a is not 0), so Eq.1 can be rewritten:

2E2 x 2E-4=(2x2)(10^2 x 10^-4)=4x10^(2-4)=4x10^-2=4/10^2=4/100=0.04

The answer is: 2E2 x 2E-4=4E-2 (=4x10^-2=0.04)

* The solutions shown above are for explanations.   You can solve this question just  multiplying coefficients, and then exponents separately, e.g. 2E3 x 6E-5
=(2x6)E(3-5)=12E-2=1.2E-1 (=1.2x10^-1=0.12)