A human hair is approximately 40 microns (um) wide. If 1 um is equal to 10
-6
meters (m),
and 1 nanometer (nm) is equal to 10-9m, then how many nanometers wide is a human hair?
A
0.0004 nm
B
0.004 nm
C
4,000 nm
D
40,000 nm

Answers

Answer 1

Answer:

D 40,000nm

Step-by-step explanation: 1nm/10^-9m * 10^-6/1um = 10^3 nm/um

Therefore in 40 microns there are:

40um * 10^3 nm/um = 40 * 10^3 nm = 40,000 nm

Answer 2

The human hair is 40,000 nanometers wide.

Option D is the correct answer.

Given,

A human hair is approximately 40 microns (um) wide.

1 um is equal to [tex]10^{-6}[/tex] meters (m)

1 nanometer (nm) is equal to [tex]10^{-9}[/tex] m.

We need to find how many nanometers wide is human hair.

How do we add exponents?

We can add the exponents if we have the same base.

Example:

a² x a³ = [tex]a^{(2+3)}[/tex] = [tex]a^{5}[/tex]

[tex]a^{-2}[/tex] x [tex]a^{-3}[/tex] = [tex]a^{(-2 +( -3))}[/tex] = [tex]a^{-5}[/tex]

We have,

Human hair = 40 microns (um) wide.

And,

1 um = [tex]10^{-6}[/tex] m

1 nm = [tex]10^{-9}[/tex] m

[tex]10^{-9}[/tex] m = 1 nm

Multiplying by 10³ on both sides.

[tex]10^{-9}[/tex] x 10³ = 10³ nm

[tex]10^{(-9+3)}[/tex] = 10³ nm

[tex]10^{-6}[/tex] m = 10³ nm

1 um = [tex]10^{-6}[/tex]m

1 um = 10³ nm

Human hair = 40 um

1 um = 10³ nm

Multiplying 40 on both sides.

40 x 1 um = 40 x 10³ nm

40 um = 40 x 1000 nm

40 um = 40,000 nm

Thus human hair is 40,000 nanometers wide.

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Answer:

STEP1:

4 Simplify — 1

Equation at the end of step1:

4 (((((((2•(a3))+(4•(a2)))-2a)-————)-(4•(a2)))-4a)+16)•(((((a2)-6a)+(4•a2))+2a)-4) (a3)

STEP2:Equation at the end of step 2

4 (((((((2•(a3))+(4•(a2)))-2a)-————)-(4•(a2)))-4a)+16)•(((((a2)-6a)+22a2)+2a)-4) (a3)

STEP 3 :

Equation at the end of step3:

4 (((((((2•(a3))+(4•(a2)))-2a)-————)-22a2)-4a)+16)•(5a2-4a-4) (a3)

STEP 4 :

4 Simplify —— a3

Equation at the end of step4:

4 (((((((2•(a3))+(4•(a2)))-2a)-——)-22a2)-4a)+16)•(5a2-4a-4) a3

STEP 5 :

Equation at the end of step5:

4 (((((((2•(a3))+22a2)-2a)-——)-22a2)-4a)+16)•(5a2-4a-4) a3

STEP 6 :

Equation at the end of step6:

4 ((((((2a3+22a2)-2a)-——)-22a2)-4a)+16)•(5a2-4a-4) a3

STEP7:Rewriting the whole as an Equivalent Fraction

 7.1   Subtracting a fraction from a whole

Rewrite the whole as a fraction using  a3  as the denominator :

2a3 + 4a2 - 2a (2a3 + 4a2 - 2a) • a3 2a3 + 4a2 - 2a = —————————————— = ————————————————————— 1 a3

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

STEP8:Pulling out like terms

 8.1     Pull out like factors :

   2a3 + 4a2 - 2a  =   2a • (a2 + 2a - 1) 

Trying to factor by splitting the middle term

 8.2     Factoring  a2 + 2a - 1 

The first term is,  a2  its coefficient is  1 .

The middle term is,  +2a  its coefficient is  2 .

The last term, "the constant", is  -1 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -1 = -1 

Step-2 : Find two factors of  -1  whose sum equals the coefficient of the middle term, which is   2 .

     -1   +   1   =   0

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Adding fractions that have a common denominator :

 8.3       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:

2a • (a2+2a-1) • a3 - (4) 2a6 + 4a5 - 2a4 - 4 ————————————————————————— = ——————————————————— a3 a3

Equation at the end of step8:

(2a6+4a5-2a4-4) (((———————————————-22a2)-4a)+16)•(5a2-4a-4) a3

STEP9:

Rewriting the whole as an Equivalent Fraction :

 9.1   Subtracting a whole from a fraction

Rewrite the whole as a fraction using  a3  as the denominator :

22a2 22a2 • a3 22a2 = ———— = ————————— 1 a3

STEP10:

Pulling out like terms :

 10.1     Pull out like factors :

   2a6 + 4a5 - 2a4 - 4  = 

  2 • (a6 + 2a5 - a4 - 2) 

Checking for a perfect cube :

 10.2    a6 + 2a5 - a4 - 2  is not a perfect cube

Trying to factor by pulling out :

 10.3      Factoring:  a6 + 2a5 - a4 - 2 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -a4 - 2 

Group 2:  a6 + 2a5 

Pull out from each group separately :

Group 1:   (a4 + 2) • (-1)

Group 2:   (a + 2) • (a5)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

 10.4    Find roots (zeroes) of :       F(a) = a6 + 2a5 - a4 - 2

Polynomial Roots Calculator is a set of methods aimed at finding values of  a  for which   F(a)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  a  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  -2.

 The factor(s) are:

of the Leading Coefficient :  1

 of the Trailing Constant :

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Answers

Answer:

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Step-by-step explanation:

Answer:

B

Step-by-step explanation:

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