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3 years ago

Find four fractions between 1/10 and 1/8

Mathematics

1 answer:

Alex17521 [72]3 years ago

6 0

9514 1404 393

Answer:

5/48, 5/46, 5/44, 5/42

Step-by-step explanation:

We can choose unit fractions with denominators between 8 and 10, separated by (10-8)/5 = 0.4 units:

1/8.4 = 5/42

1/8.8 = 5/44

1/9.2 = 5/46

1/9.6 = 5/48

<em>Check</em>

1/8 = 0.125
5/42 ≈ 0.119
5/44 ≈ 0.114
5/46 ≈ 0.109
5/48 ≈ 0.104
1/10 = 0.100

_____

<em>Additional comment</em>

There are an infinite number of such fractions. We are given unit fractions with different denominators, so it works reasonably well to choose denominators between those given. Then the trick is to convert the fraction to a ratio of integers. In this case, multiplying by (5/5) does the trick.

Another approach is to write the fractions with a common denominator, then choose numerators between the ones given. For example, 1/10 = 4/40, and 1/8 = 5/40, so you could write some fractions with numerators between 4 and 5. Possibilities are 4.1/40 = 41/400, 4.3/40 = 43/400, 4.7/40 = 47/400, 4.9/40 = 49/400.

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2 years ago

the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

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Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$

$\frac{db}{dt} = -2.262\,\frac{cm}{min}$

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

6 0

3 years ago