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Oksi-84 [34.3K]
3 years ago
12

22. Write the following as fractions: i) Seven hundredths ii) Six ninths​

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
3 0
Here are both in fraction form.

i) 7/100
ii) 6/9
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Find x: 3/(x-4)(x-7) + 6/(x-7)(x-13) + 15/(x-13)(x-28) - 1/x-28 = -1/20
Novay_Z [31]

The value of x<em> </em>in the polynomial fraction 3/((x-4)•(x-7)) + 6/((x-7)•(x-13)) + 15/((x-13)•(x-28)) - 1/(x-28) = -1/20 is <em>x </em>= 24

<h3>How can the polynomial with fractions be simplified to find<em> </em><em>x</em>?</h3>

The given equation is presented as follows;

\frac{3}{(x - 4) \cdot (x - 7) }  + \frac{6}{(x - 7) \cdot (x - 13)   }  + +\frac{15}{(x - 13) \cdot (x - 28) } - \frac{1}{(x - 28)  } =  -  \frac{1}{20}

Factoring the common denominator, we have;

\frac{3\cdot(x - 13) \cdot(x - 28) + 6 \cdot(x - 4) \cdot(x - 28)  + 15 \cdot(x - 4) \cdot(x - 7)  - (x - 4) \cdot (x - 7)\cdot(x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)}   + =  -  \frac{1}{20}

Simplifying the numerator of the right hand side using a graphing calculator, we get;

By expanding and collecting, the terms of the numerator gives;

-(x³ - 48•x + 651•x - 2548)

Given that the terms of the numerator have several factors in common, we get;

-(x³ - 48•x + 651•x - 2548) = -(x-7)•(x-28)•(x-13)

Which gives;

\frac{-(x - 7) \cdot(x - 28)\cdot (x - 13)}{(x - 4) \cdot (x - 7)\cdot(x - 13) \cdot(x - 28)}   + =  -  \frac{1}{20}

Which gives;

\frac{-1}{(x - 4)}   + =  -  \frac{1}{20}

x - 4 = 20

Therefore;

  • x = 20 + 4 = 24

Learn more about polynomials with fractions here:

brainly.com/question/12262414

#SPJ1

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2 years ago
( Will hand out brainiest) A California condor left its cliff-side nest and soared downwards at a rate of 16 meters per second.
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First calculate the distance covered going down:

d_down = (16 m / s) * 8 s = 128 m

 

Then the distance going up is:

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So the distance from the ledge to the nest is:

d = 128 m – 71 m = 57 m

 

Therefore the elevation is:

<span>elevation = 1364 m + 57 m = 1421 m</span>

4 0
3 years ago
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