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3 years ago

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A hockey player strikes a hockey puck. The height of the puck increases until it reaches a maximum height of 3 feet, 55 feet awa

y from the player. The height $y$ (in feet) of a second hockey puck is modeled by $y=x\left(0.15-0.001x\right)$ , where $x$ is the horizontal distance (in feet). Compare the distances traveled by the hockey pucks before hitting the ground.

1 answer:

tatuchka [14]3 years ago

5 0

Answer:

Second puck travels farther

Step-by-step explanation:

Maximum height of first puck = 3 feet

The height of a second hockey puck is modeled by:

$y=x\left(0.15-0.001x\right)$

$y=0.15x-0.001x^2$

To find maximum height of second puck

$y'=0.15-0.002x$

Equate the derivative equals to 0

0.15-0.002x=0

$\frac{0.15}{0.002}=x$

75 = x

At x = 75

$y=0.15(75)-0.001(75)^2=5.625$

So, The maximum height of second puck is greater than first puck

So, Second puck travels farther

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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

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Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

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$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

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By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

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But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

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$a$ is an arbitrary prime number. (1)

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$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

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Answer:

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Step-by-step explanation:

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