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yuradex [85]
2 years ago
14

The edges of a shoebox are measured to be 11.3 cm, 18.9 cm, and 29 cm. Determine the volume of the box retaining the proper numb

er of significant figures in your answer.
Mathematics
1 answer:
puteri [66]2 years ago
7 0

The volume (V) is given in terms of length (L), width (W), and height (H) by the formula ...

... V = LWH

Substituting your dimensions, you have

... V = (11.3 cm)·(18.9 cm)·(29 cm) = 6193.53 cm³ ≈ 6200 cm³

_____

The factor 29 cm has 2 significant digits, so that is the precision required of the answer. It could also be written with less ambiguity as to precision as 6.2 L. (Trailing zeros to the left of the decimal point are always ambiguous when it comes to significant digits.)

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Step-by-step explanation:

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Evaluate the function at the given numbers( correct to six decimal places). Use the results to guess the value of the limit or e
netineya [11]

Answer:

Value of the limit is 0.5.

Step-by-step explanation:

Given,

F(x)=\frac{e^x-1-x}{x^2}

When,

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x=0.5, F(0.5)=\frac{e^0.5-1-0.5}{(0.5)^2}=0.594885

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x=0.05, F(0.05)=\frac{e^0.05-1-0.05}{(0.05)^2}=0.508438

x=0.01, F(0.01)=\frac{e^0.01-1-0.01}{(0.01)^2}=0.501670 \hfill (1)

Correct upto six decimal places.

Now,

\lim_{x\to 0}F(x)=\lim_{x\to 0}\frac{e^x-1-x}{x^2}   (\frac{0}{0}) form, applying L-Hospital rule that is differentiating numerator and denominator we get,

\lim_{x\to 0}F(x)

=\lim_{x\to 0}\frac{e^x-1}{2x}    (\frac{0}{0}) form.

=\lim_{x\to 0}\frac{e^x}{2}=\frac{1}{2}=0.5\hfill (2)

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What’s the value of x? <br> y= 5x + 9<br> y= -x + 3
ser-zykov [4K]

Answer:

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Step-by-step explanation:

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For any two documents x and z, define k(x; z) to equal the number of unique words that occur in both x and z (i.e., the size of
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Answer:

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Given

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This gives;

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