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Sergeu [11.5K]
3 years ago
10

What is 12/24, 3/30, 10/8, 34/20, and 14/8 in lowest term and in percent form

Mathematics
1 answer:
kirza4 [7]3 years ago
3 0
To find a fraction's lowest term, just simplify them and divide to find the decimal form.
12/24 = 1/2, 0.5
3/30 = 1/10, 0.1
10/8 = 5/4, 1.25
34/20 = 17/10, 1.7
14/8 = 7/4, 1.75

Finished!
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What is the difference?
Crazy boy [7]

Answer:

<h2>D. \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\ or StartFraction x squared + 3 x minus 12 Over (x + 3) (x minus 5) (x + 7) EndFraction</h2>

Step-by-step explanation:

Given the expression \frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }, the dfference is expressed as follows;

Step1: First we need to factorize the denominator of each function.

\frac{x}{x^{2}-2x-15 } - \frac{4}{x^{2} + 2x - 35 }\\= \frac{x}{x^{2}-5x+3x-15 } - \frac{4}{x^{2} + 7x-5x - 35 }\\= \frac{x}{x(x-5)+3(x-5) } - \frac{4}{x( x+ 7)x-5(x +7) }\\= \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\\\

Step 2: We will find the LCM of the resulting expression

=  \frac{x}{(x-5)(x+3) } - \frac{4}{(x-5)(x +7) }\\= \frac{x(x+7)-4(x+3)}{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+7x-4x-12 }{(x-5)(x+3)(x+7)} \\= \frac{x^{2}+3x-12 }{(x-5)(x+3)(x+7)} \\

The final expression gives the difference

6 0
3 years ago
Take a factor out of the square root:
Nat2105 [25]

Answer:

See below.

Step-by-step explanation:

Lets take √72:

Factoring 72 to find perfect squares:

72 = 2*2*2*3*3

The largest perfect square in the above = 2*2*3*3 = 36.

So we can write 72 as 36*2.

√72 = √(36 * 2)

√36 * √2

= 6√2.

One more example:

√18  = √(2*3*3)

= √9*√2

= 3√2.

6 0
3 years ago
The lengths of a triangle are given below is the triangle obtuse acute or right ? 11,12,15
andriy [413]
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6 0
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Read 2 more answers
Write 2.9302 as a fraction and a mixed number​
bonufazy [111]

rationalise 29302/100000

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8 0
3 years ago
Han can run 100 meters in 20 seconds how long will it take him to run 3000 metres
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Answer:

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Step-by-step explanation:

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Next to get our answer we have to calculate the amount of time

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3 years ago
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