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Thepotemich [5.8K]
3 years ago
11

Solving Rational equations. LCD method. Show work. Image attached.

Mathematics
1 answer:
Ugo [173]3 years ago
5 0

Answer:

k=\frac{1}{2}

Step-by-step explanation:

The given rational equation is

\frac{5k}{k+2}+\frac{2}{k}=5

The Least Common Denominator is

k(k+2)

Multiply each term by the LCD.

k(k+2)\times \frac{5k}{k+2}+k(k+2)\times \frac{2}{k}=5k(k+2)

Simplify;

k\times \frac{5k}{1}+(k+2)\times \frac{2}{1}=5k(k+2)

\Rightarrow k(5k)+2(k+2)=5k(k+2)

Expand;

\Rightarrow 5k^2+2k+4=5k^2+10k

Group similar terms;

\Rightarrow 5k^2-5k^2+2k-10k=-4

\Rightarrow -8k=-4

Divide by -8.

k=\frac{1}{2}

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The ratio of cats to all animals is 12:23.
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3 years ago
Please help me out with this
Genrish500 [490]

Answer:

x = 17

Step-by-step explanation:

For the parallelogram to be a rhombus then then the diagonal must bisect the given angle, thus

3x - 11 = x + 23 ( subtract x from both sides )

2x - 11 = 23 ( add 1 to both sides )

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7 0
3 years ago
What is the value of x
djyliett [7]

Hey there!


Assuming #1 ….


x + 7x + x = 180°
REWORD

1x + 7x + 1x = 180°

COMBINE the LIKE TERMS

(1x + 7x + 1x) = 180°

8x + 1x = 180°

9x = 180°

DIVIDE 9 to BOTH SIDES

9x/9 = 180/9

CANCEL out: 9/9 because it equal 1

KEEP: 180/9 because it give you the x-value

NEW EQUATION: x = 180/9

SIMPLIFY IT!
x = 20
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Good luck on your assignment & enjoy your day!


~Amphitrite1040:)

3 0
2 years ago
Read 2 more answers
Is 8.3 greater than 83 tenths?
kati45 [8]

Answer:

<em>NO</em>

Step-by-step explanation:

They are equal

83 tenths divided by 10 is 8.3

<u>Hope this helps :-)</u>

8 0
3 years ago
The sum of twice a number and 13 is 75
Nina [5.8K]
So what they are saying is the sum of 2x +13=75
Therefore we can just solve for 2x, our unknown value.
2x+13=75 now subtract 13 and divide by 2 from both sides of equation to not change the equation
x+13-13=75-13
This leaves us with
x=(75-13)/2=31 therefore x=31 or our unknown value is 31.

I hope this helped. Any questions please just ask. Thank you.
3 0
3 years ago
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