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Yuri [45]
3 years ago
14

Simplyfy (x - 5 / x^3 + 27) + (2 / x^2 - 9)

Mathematics
2 answers:
BARSIC [14]3 years ago
5 0
\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\ \quad \\
\textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 \\\\
-------------------------------\\\\

\bf \cfrac{x-5}{x^3+27}+\cfrac{2}{x^2-9}\implies \cfrac{x-5}{x^3+3^3}+\cfrac{2}{x^2-3^2}
\\\\\\
\cfrac{x-5}{(x+3)(x^2-3x+3^2)}+\cfrac{2}{(x-3)(x+3)}\\\\\\
\cfrac{x-5}{(x+3)(x^2-3x+9)}+\cfrac{2}{(x-3)(x+3)}
\\\\\\
\textit{so, we'll use the LCD of }(x-3)(x+3)(x^2-3x+9)
\\\\\\
\cfrac{(x-3)(x-5)~~+~~(x^2-3x+9)2}{(x-3)(x+3)(x^2-3x+9)}
\\\\\\
\cfrac{x^2-8x+15~~+~~2x^2-6x+18}{(x-3)(x+3)(x^2-3x+9)}\implies \cfrac{3x^2-14x+33}{(x-3)(x+3)(x^2-3x+9)}
Maksim231197 [3]3 years ago
3 0
The answer simplified should be x-5/x^3+18+2/x^2. Hope that is helpful.
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cricket20 [7]

Answer:

Largest Median: Same

Largest Range: Castro

Largest IQR: Castro

Step-by-step explanation:

With a box-and-whisker plot, the box represents the upper and lower quartiles, the vertical line inside the box represents the median, and the lines on either side of the box show the high and low of the range.

Largest Median: Medians are the same because the verticle line inside the boxes is at 7 for both

Largest Range: Ms Castro's Class- the lines on either side of the box for Ms Castro go from 1-10 while the other class only goes from 4-10.

Largest IQR (interquartile range) Ms. Castro's class: their IQR goes from 5-8 while the other class only goes from 6-8

6 0
3 years ago
Given here are a set of sample data: 12.0, 18.3, 29.6, 14.3, and 27.8. the sample standard deviation for this data is equal to _
disa [49]
The answer is 7.9306

Using the formula in the attached:
Where: xi = sample value; μ = sample mean; n = sample size

1.) Calculate the mean first:
μ = 12.0 + 18.3 + 29.6 + 14.3 + 27.8 / 5
   = 102 / 5
μ = 20.4

2.) Using the mean, calculate (xi - μ)² for each value:
(12.0 - 20.4)² = 70.56
(18.3 - 20.4)² = 4.41
(29.6 - 20.4)² = 84.64
(14.3 - 20.4)² = 37.21
(27.8 - 20.4)² = 54.76

3.) Sum the squared differences and divide by n - 1.
μ = 70.56 + 4.41 + 84.64 + 37.21 + 54.76
   = 251.58 / 5-1
μ = 62.895 (this is now called sample variance)

4.) Get the square root of the sample variance:
 √62.895 = 7.9306

3 0
3 years ago
the area of the rectangle 56cm. the length is 2cm more than x and the with is 5cm less than twice x. solve for x. round the near
sattari [20]
(x+2)(2x-5)=56

2x^2-5x+4x-10=56

2x^2-x-10=56

2x^2-x-66=0

2x^2-12x+11x-66=0

2x(x-6)+11(x-6)=0

(2x+11)(x-6)=0, since x is a measurement x>0 so:

x=6cm


7 0
3 years ago
Read 2 more answers
If 3.N=1,then N is the
rosijanka [135]
3 * N = 1

The multiplicative inverse is just the reciprocal of the number. The reciprocal is just " flipping " the number. So the reciprocal of 3 (or 3/1) is 1/3. And any time u multiply a number by its multiplicative inverse (reciprocal), the result is 1

3 * 1/3 = 3/3 = 1.....so N is ur multiplicative inverse
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VladimirAG [237]
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