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3 years ago

Write the number below as a fraction in its simplest form 0.673 the 7 and 3 are recurring

Mathematics

1 answer:

wlad13 [49]3 years ago

7 0

Let x=0.6737373...(1)
10x=6.737373...(2)
1000 x=673.737373...(3)
(3)-(2) gives
990x=667
x=667/990

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Write 75% as a decimal

aivan3 [116]

Answer:

0.75

Step-by-step explanation:

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3 years ago

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If p, q are zeroes of polynomial f(x) =2x^2-7x+3,find the value of p^2+q^2

Elza [17]

Answer:

p^2 + q^2 = 22/9.

Step-by-step explanation:

For a quadratic function ax^2 + bx + c if the zeroes are A and B then

A + B = -b/a and AB = c/a.

2x^2 - 7x + 3

Now p and q are the zeroes of the above function so

p + q = 7/3 and pq = 3/2.

Now p^2 + q^2 = (p + q)^2 - 2pq

= (7/3)^2 - 2* 3/2

= 49/9 - 3

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= 22/9.

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2 years ago

Find the area of this parallelogram. 8.9 cm 7.1 cm 11.4 cm

defon

<h3>Answer: 80.94 square cm</h3>

==================================================

Work Shown:

Area of parallelogram = base*height

Area of parallelogram = 11.4*7.1

Area of parallelogram = 80.94 square cm

-------------

notes:

The base and height always form a 90 degree angle (as shown by the small square marker).
The side length of 8.9 cm is never used.
We can write "square cm" as "cm^2" or $\text{ cm}^2$

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3 years ago

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One month Leila rented 3 movies and 2 video games for a total of $21 . The next month she rented 8 movies and 4 video games for

lina2011 [118]

x = price movie = 4.25
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3 years ago

If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

and so (a + c)/(b + d) > a/b.

Then together we get the desired inequality.

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3 years ago