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DaniilM [7]
4 years ago
14

David measured the height of a window frame as 5.61 feet, but the actual height was 6 feet. What is the percentage of error in D

avid's measurement?
Mathematics
1 answer:
Nataly [62]4 years ago
4 0
<h2>Percentage of error in David's measurement is 6.5 %</h2>

Step-by-step explanation:

Measured height = 5.61 feet

Original height = 6 feet

Error in measurement = Original height - Measured height

Error in measurement = 6 - 5.61

Error in measurement = 0.39 feet

\texttt{Percentage of error in David's measurement = }\frac{\texttt{Error in measurement}}{\texttt{Original height}}\times 100\\\\\texttt{Percentage of error in David's measurement = }\frac{0.39}{6}\times 100\\\\\texttt{Percentage of error in David's measurement = }6.5\%

Percentage of error in David's measurement is 6.5 %

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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.

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⇔ 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

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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

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We have:

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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

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Answer:

see explanation

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D (2, - 4 ) → D' (4, 2 )

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