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GalinKa [24]
3 years ago
15

Convert 1.50 micrometers^2 to m^2. Show your solution.

Mathematics
1 answer:
Vadim26 [7]3 years ago
8 0

Here is the solution for the problem:

<span>1m * 1m = 1m² </span><span>
<span>1m = 1,000,000µm = 1 x 10^6µm </span>
<span>1 x 10^6µm * 1 x 10^6µm = 1m² </span>
<span>1x10^12µm² = 1m² (Now, divide both sides by 1x10^12). </span>
<span>1x10^12µm² / (1 x 10^12) = 1m² / (1 x 10^12) </span>
<span>1µm² = 0∙000 000 000 001 m² (x both sides by 1.5). </span>
<span>1.5µm² = 0∙000 000 000 0015 m² </span>
1.5µm² = 1∙5x10^-12 m²</span>

So 1.50 micrometers^2 is equal to 1.5 x 10^-12 m². I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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Compare the values 4x107 is what times as much as 4x103
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Answer:

10^{4}

Step-by-step explanation:

4 x 10^7 is 40000000

4 x 10^3 is 4000

We are told that 4x10^7 is something times as much as 4x10^3

So we will work backwards to find the answer,

If you divide 40000000 by 4000 you will get 10000

10000 is the same as 10^4

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3 years ago
1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.
german

Answer:

Step-by-step explanation:

1.

To write the form of the partial fraction decomposition of the rational expression:

We have:

\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}

2.

Using partial fraction decomposition to find the definite integral of:

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx

By using the long division method; we have:

x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }

                  - 2x^3 -16x^2-40x

                 <u>                                         </u>

                                            x+ 20

So;

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}

By using partial fraction decomposition:

\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}

                         = \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now;  we have to relate like terms on both sides; we have:

A + B = 1   ;   2A - 10 B = 20

By solvong the expressions above; we have:

A = \dfrac{5}{2}     B =  \dfrac{3}{2}

Now;

\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}

Thus;

\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}

Now; the integral is:

\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx =  \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx

\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx =  x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0
3 years ago
Can you please help?
Liula [17]

Answer:

\huge\boxed{3.873}

Step-by-step explanation:

Note that the squares are on the sides of these triangles are on the hypotenuse/one leg.

We know that the area of a square is l^2 where l is the length. Since we know the area, we can find the square root of the area to find the side length.

So the side lengths are \sqrt{25} and \sqrt{10}.

Since we know the hypotenuse is \sqrt{25} and one of the legs is \sqrt{10}, we can use the Pythagorean Theorem to find the missing side.

The Pythagorean Theorem states that a^2 + b^2 = c^2, where c is the hypotenuse and a/b are the legs.

We know the <em>hypotenuse </em>and <em>a leg</em>, so we can substitute inside

a^2 + \sqrt{10}^2 = \sqrt{25}^2

Squaring a square root is the same as doing nothing.

a^2 + 10 = 25\\\\a^2 = 25-10\\\\a^2 = 15\\\\a = \sqrt{15}\\\\a \approx 3.873

Hope this helped!

7 0
3 years ago
Simplify {(1/3)^-2 -(1/2)^-2 } / (1/4)^-2
Mumz [18]

Answer: 209/16 = 13.06250

Step-by-step explanation: hope it helps

6 0
3 years ago
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