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2 years ago

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What is the difference? startfraction 2 x 5 over x squared minus 3 x endfraction minus startfraction 3 x 5 over x cubed minus 9

x endfraction minus startfraction x 1 over x squared minus 9 endfraction startfraction (x 5) (x 2) over x cubed minus 9 x endfraction startfraction (x 5) (x 4) over x cubed minus 9 x endfraction startfraction negative 2 x 11 over x cubed minus 12 x minus 9 endfraction startfraction 3 (x 2) over x squared minus 3 x endfraction

1 answer:

weeeeeb [17]2 years ago

6 0

The difference of the expression $\frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x}$ is $\frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}$

<h3>How to determine the difference?</h3>

The expression is given as:

$\frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x}$

Factor the denominators of the expressions:

$\frac{2x^5}{x(x - 3)} - \frac{3x^5}{x(x^2 - 9)}$

Apply the difference of two squares to x² - 9

$\frac{2x^5}{x(x - 3)} - \frac{3x^5}{x(x - 3)(x + 3)}$

Take LCM

$\frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}$

Hence, the difference of the expression $\frac{2x^5}{x^2 - 3x} - \frac{3x^5}{x^3 - 9x}$ is $\frac{2x^5(x + 3) - 3x^5}{x(x - 3)(x + 3)}$

Read more about expressions at:

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Answer:

Part a) The new rectangle labeled in the attached figure N 2

Part b) The diagram of the new rectangle with their areas in the attached figure N 3, and the trinomial is $x^{2} +11x+28$

Part c) The area of the second rectangle is 54 in^2

Part d) see the explanation

Step-by-step explanation:

The complete question in the attached figure N 1

Part a) If the original square is shown below with side lengths marked with x, label the second diagram to represent the new rectangle constructed by increasing the sides as described above

we know that

The dimensions of the new rectangle will be

$Length=(x+4)\ in$

$width=(x+7)\ in$

The diagram of the new rectangle in the attached figure N 2

Part b) Label each portion of the second diagram with their areas in terms of x (when applicable) State the product of (x+4) and (x+7) as a trinomial

The diagram of the new rectangle with their areas in the attached figure N 3

we have that

To find out the area of each portion, multiply its length by its width

$A1=(x)(x)=x^{2}\ in^2$

$A2=(4)(x)=4x\ in^2$

$A3=(x)(7)=7x\ in^2$

$A4=(4)(7)=28\ in^2$

The total area of the second rectangle is the sum of the four areas

$A=A1+A2+A3+A4$

State the product of (x+4) and (x+7) as a trinomial

$(x+4)(x+7)=x^{2}+7x+4x+28=x^{2} +11x+28$

Part c) If the original square had a side length of x = 2 inches, then what is the area of the second rectangle?

we know that

The area of the second rectangle is equal to

$A=A1+A2+A3+A4$

For x=2 in

substitute the value of x in the area of each portion

$A1=(2)(2)=4\ in^2$

$A2=(4)(2)=8\ in^2$

$A3=(2)(7)=14\ in^2$

$A4=(4)(7)=28\ in^2$

$A=4+8+14+28$

$A=54\ in^2$

Part d) Verify that the trinomial you found in Part b) has the same value as Part c) for x=2 in

We have that

The trinomial is

$A(x)=x^{2} +11x+28$

For x=2 in

substitute and solve for A(x)

$A(2)=2^{2} +11(2)+28$

$A(2)=4 +22+28$

$A(2)=54\ in^2$ ----> verified

therefore

The trinomial represent the total area of the second rectangle

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4 years ago

SCORPION-xisa [38]

53 minutes

3:08 to 4:00 is 52 minutes.

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3 years ago

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Inessa05 [86]

Answer:

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Step-by-step explanation:

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tigry1 [53]

Answer:

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Step-by-step explanation:

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now subtract n from 4n and subtract 4 from 6

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Since we have given that

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Here, we use "Binomial distribution":

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