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Jet001 [13]
3 years ago
9

An ultramarathon relay of 37.2

Mathematics
1 answer:
Naya [18.7K]3 years ago
6 0

Equation represent situation is 6d = 37.2

<u>Given:</u>

Total distance cover by runners = 37.2 miles

Number of runners = 6

<u>Find:</u>

Equation represent situation

<u>Computation:</u>

Assume;

Each runner cover distance = d miles

So.

Total distance cover by runners = Number of runners × Each runner cover distance

37.2 = 6 × d

6d = 37.2

Learn more:

brainly.com/question/15172156?referrer=searchResults

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What is the difference?
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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
The amount of radioactive element remaining, r, in a 100mg sample after d days is represented using the equation r=100(1/2) d/5.
Ludmilka [50]

Answer:

   12.94%

Step-by-step explanation:

r = 100(1/2)^(d/5) = 100((1/2)^(1/5))^d ≈ 100(.87055)^d

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4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cfrac%20%7B%2018%20%5E%20%7B%20n%20%2B%201%20%7D%20%5Ctimes%203%20%5E%20%7B%201%20-%20n%20%7
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Answer:

4 \times  {3}^{n + 3}

Step-by-step explanation:

\frac{ {18}^{n + 1}  \times {3}^{1 - n}}{  {2}^{n - 1}  }

\frac{ {3}^{2n + 2}  \times  {2}^{n + 1}  \times  {3}^{1 - n} }{ {2}^{n - 1} }

{3}^{2n + 2}  \times  {2}^{2}  \times  {3}^{1 - n}

{3}^{2n + 2}  \times 4 \times  {3}^{1 - n}

= 4 \times  {3}^{n + 3}

8 0
3 years ago
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Answer:

IM pretty sure b) and c)

Step-by-step explanation:

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Step-by-step explanation: I searched up ways to learn abc math and it said guess so yea

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