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daser333 [38]
4 years ago
11

How complex fractions can be used with ratios

Mathematics
1 answer:
STatiana [176]4 years ago
3 0
Complex fractions are numbers in which the numerator and the denominator are both fractions. in this case, to solve the ratio, we can multiply the numerator fraction by the reciprocal of the denominator fraction. Another way is to solve the fraction separately and then divide eventually. 
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Step-by-step explanation:

315 centimeters converted to meters is 3.15 meters.

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What are the domain and range of the function f(x)=3 over 4 x+5
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Step-by-step explanation:

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3 years ago
1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is
sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by d(t)=\frac{t^{2}}{6}+4t where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}

so let's take the first interval:

(a) [1.95, 2.05]

V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}

we get that:

d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125

d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167

so:

V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s

(b) [1.995, 2.005]

V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}

we get that:

d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831

d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835

so:

V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s

(c) [1.9995, 2.0005]

V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}

we get that:

d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358

d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358

so:

V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s

(d) [2, 2.00001]

V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}

we get that:

d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333

d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333

so:

V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0
3 years ago
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