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Karo-lina-s [1.5K]
3 years ago
15

(a) What is 0.36 repeating 36 expressed as a fraction in simplest form? (b) What is 0.36 repeating 6 expressed as a fraction in

simplest form? (c) What is expressed 0.36 as a fraction in simplest form?
Please help
Mathematics
2 answers:
sweet [91]3 years ago
6 0
A. 4/11
b. 11/30
c. 9/25
GREYUIT [131]3 years ago
4 0

Answer:

a) x=\frac{4}{11}    

b) x=\frac{11}{30}  

c) x=\frac{9}{25}  

Step-by-step explanation:

(a) Given : 0.36 repeating 36

To find : What is 0.36 repeating 36 expressed as a fraction in simplest form?

Solution :

Let x=0.36363636363.....   ......(1)

Now as there are two digits repeating immediately after decimal point, we multiply above by  100  and get,

100x=36.36363636....  .....(2)

Subtract equation (1) from (2),

99x=36

x=\frac{36}{99}

x=\frac{4}{11}    

(b) Given : 0.36 repeating 6

To find : What is 0.36 repeating 6 expressed as a fraction in simplest form?

Solution :

Let x=0.3666666.....   ......(1)

Now as there are one digits repeating immediately after decimal point, we multiply above by  10  and get,

10x=3.6666666....  .....(2)

Subtract equation (1) from (2),

9x=3.3

x=\frac{3.3}{9}

x=\frac{11}{30}  

(c) Given : 0.36

To find : What is expressed 0.36 as a fraction in simplest form?

Solution :

Let x=0.36

x=\frac{36}{100}

x=\frac{9}{25}  

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Express 2x+1/(x-2)(x²+1) as a partial fraction.
Oduvanchick [21]

Answer:

Partial fraction = 1/(x-2) - x/(x^2+1)

Step-by-step explanation:

Question:

Express 2x+1/(x-2)(x²+1) as a partial fraction.

Note: it will be assumed that there was a typo in the interpretation of parentheses to mean

(2x+1) / ( (x-2)(x^2+1) )

Let

(2x+1) / ( (x-2)(x^2+1) ) = A/(x-2) + (Bx+C)/(x^2+1) .........................(0)

(2x+1) / ( (x-2)(x^2+1) ) = (A(x^2+1)+(Bx+C)(x-2)) / ( (x-2)(B/(x^2+1) )

(2x+1) / ( (x-2)(x^2+1) ) = (Ax^2+A+Bx^2+(C-2B)x-2C) / ( (x-2)(B/(x^2+1) )

(2x+1) / ( (x-2)(x^2+1) ) = ( (A+B)x^2+(C-2B)x+A-2C ) / ( (x-2)(B/(x^2+1) )

Match numerators

2x+1 = (A+B)x^2+(C-2B)x+A-2C

Match coefficients,

A+B = 0 ..................(1)

-2B+C = 2 .................(2)

A-2C = 1 ...................(3)

Solve for A, B and C

Substitute A from (1) in (3)

-B - 2C =1  

transpose and solve for B

B = -2C-1  ....................(4)

Substitue B from (4) in (2)

-2(-2C-1) + C = 2  

simplify

5C = 2-2 = 0

C=0  ..........................(5)

substitute (5)  in (4)

B = -2C-1 = -1  ...............(6)

Substitue (6) in (1)

A+(-1) = 0

A=1 .............................(7)

Using values from (7), (6) and (5) to substitute in (0)

we get

(2x+1) / ( (x-2)(x^2+1) ) = 1/(x-2) - x/(x^2+1)

as the required partial fraction

7 0
3 years ago
Read 2 more answers
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