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3 years ago

Which pair of numbers is relatively prime? A. 42 and 77 B. 34 and 55 C. 45 and 102 D. 99 and 123

Mathematics

1 answer:

leonid [27]3 years ago

8 0

Two pair of numbers are said to be relatively prime if there is no integer greater than 1, that divides them both.

Consider the given pair of the numbers to identify which pair is relatively prime.

1. Consider 42 and 77

7 is the number which divides 42 and 77 both. Therefore, they are not relatively prime.

2. Consider 34 and 55

Since, there is no number greater than 1, which divides both the numbers. So, they are relatively prime numbers.

3. Consider 45 and 102

3 is the number divides 45 and 102 both. Therefore, they are not relatively prime.

4. Consider 99 and 123

3 is the number divides 99 and 123 both. Therefore, they are not relatively prime.

Therefore, Option B is the correct answer.

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nordsb [41]

Answer:

$h'(t) = -12.6 t +53$

Now we can set up the derivate equal to 0 and we have:

$-12.6 t +53 = 0$

And solving for t we got:

$t = \frac{53}{12.6}= 4.206$

For the second derivate respect the time we got:

$h''(t) = -12.6$

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

$h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft$

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

Step-by-step explanation:

For this case we have the following function:

$h(t) = -6.3t^2 +53 t+24$

In order to maximize this function we need to take the first derivate respect the time and we have:

$h'(t) = -12.6 t +53$

Now we can set up the derivate equal to 0 and we have:

$-12.6 t +53 = 0$

And solving for t we got:

$t = \frac{53}{12.6}= 4.206$

For the second derivate respect the time we got:

$h''(t) = -12.6$

So then we can conclude that t = 4.206 is a maximum for the function.

And the corresponding height would be:

$h(t=4.206) = -6.3(4.206)^2 +53*4.206+24= 135.468 ft$

So the maximum occurs at t = 4.206 s and with a height of 135.468 m

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