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Korolek [52]
1 year ago
9

Study the following data set.

Mathematics
1 answer:
Rudiy271 year ago
5 0

Answer:16.1

Step-by-step explanation:

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Scientific notation for 3427
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3.427 *10^3

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the 10 . If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

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3 years ago
Will Mark Brainiest!!! Simplify the following:
xz_007 [3.2K]

Answer:

1.B

2.A

3. B

Step-by-step explanation:

1. \frac{x+5}{x^{2} + 6x +5 }

We have the denominator of the fraction as following:

x^{2} + 6x + 5 \\= x^{2} + (1 + 5)x + 5\\= x*x + 1x + 5x + 5*1\\= x ( x + 1) + 5(x + 1)\\= (x + 1) (x + 5)

As the initial one is a fraction, so that its denominator has to be different from 0.

=> (x^{2} +6x+5) ≠ 0

⇔ (x +1) (x +5) ≠ 0

⇔ (x + 1) ≠ 0; (x +5) ≠ 0

⇔ x ≠ -1; x ≠ -5

Replace it into the initial equation, we have:

\frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)}

As (x+5) ≠ 0; we divide both numerator and denominator of the fraction by (x +5)

=> \frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)} = \frac{1}{x+1}

So that \frac{x+5}{x^{2} + 6x +5 } = \frac{1}{x+1} with x ≠ 1; x ≠ -5

So that the answer is B.

2. \frac{(\frac{x^{2} -16 }{x-1} )}{x+4}

As the initial one is a fraction, so that its denominator has to be different from 0

=> x + 4 ≠ 0

=> x ≠ -4

As \frac{x^{2}-16 }{x-1} is also a fraction, so that its denominator (x-1) has to be different from 0

=> x - 1 ≠ 0

=> x ≠ 1

We have an equation: x^{2} - y^{2} = (x - y ) (x+y)

=> x^{2} - 16 = x^{2} - 4^{2} = (x -4)  (x +4)

Replace it into the initial equation, we have:

\frac{(\frac{x^{2} -16 }{x-1} )}{x+4} \\= \frac{x^{2} -16 }{x-1} . \frac{1}{x + 4}\\= \frac{(x-4)(x+4)}{x-1}. \frac{1}{x + 4}

As (x + 4) ≠ 0 (proven above), we can divide both numerator and the denominator of the fraction by (x +4)

=> \frac{(x-4)(x+4)}{x-1} .\frac{1}{x+4} =\frac{x-4}{x-1}

So that the initial equation is equal to \frac{x-4}{x-1} with x ≠-4; x ≠1

=> So that the correct answer is A

3. \frac{x}{4x + x^{2} }

As the initial one is a fraction, so that its denominator (4x + x^2) has to be different from 0

We have:

(4x + x^2) = 4x + x.x = x ( x + 4)

So that:  (4x + x^2) ≠ 0 ⇔ x ( x + 4 ) ≠ 0

⇔ \left \{ {{x\neq 0} \atop {(x+4)\neq0 }} \right.  ⇔ \left \{ {{x\neq 0} \atop {x \neq -4 }} \right.

As (4x + x^2) = x ( x + 4) , we replace this into the initial fraction and have:

\frac{x}{4x + x^{2} } = \frac{x}{x(x+4)}

As x ≠ 0, we can divide both numerator and denominator of the fraction by x and have:

\frac{x}{x(x+4)} =\frac{x/x}{x(x+4)/x} = \frac{1}{x+4}

So that \frac{x}{4x+x^{2} }  = \frac{1}{x+4} with x ≠ 0; x ≠ -4

=> The correct answer is B

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