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yawa3891 [41]
3 years ago
14

. A hawk flying at a height of 60 feet spots a rabbit on the ground. If the hawk dives at a speed of 55 feet per second, how

Mathematics
1 answer:
Vladimir [108]3 years ago
8 0

Answer:

The hawk reaches the rabbit at t=4.3 sec

Step-by-step explanation:

Let

h ----> is the height in feet

t ----> is the time in seconds

v ---> the initial velocity in feet pr second

s ----- is the starting height

we have

h=-16t^{2} +vt+s

when the hawk reaches the rabbit the value of h is equal to zero

we have

v=55\ ft/sec

s=60\ ft

substitute

0=-16t^{2} +55t+60

Solve the quadratic equation by graphing

The solution is t=4.3 sec

see the attached figure

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The sum of the squares of two numbers is 8 . The product of the two numbers is 4 . Find the numbers.
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Hello there.

First, assume the numbers x,~y such that they satisties both affirmations:

  • The sum of the squares of two numbers is 8.
  • The product of the two numbers is 4.

With these informations, we can set the following equations:

\begin{center}\align x^2+y^2=8\\ x\cdot y=4\\\end{center}

Multiply both sides of the second equation by a factor of 2:

2\cdot x\cdot y = 2\cdot 4\\\\\\ 2xy=8~~~~~(2)^{\ast}

Make (1)-(2)^{\ast}

x^2+y^2-2xy=8-8\\\\\\ x^2-2xy+y^2=0

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(x-y)^2=0

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Substituting that information from (3) in (2), we get:

x\cdot x = 4\\\\\\ x^2=4

Calculate the square root on both sides of the equation:

\sqrt{x^2}=\sqrt{4}\\\\\\ |x|=2\\\\\\ x=\pm~2

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y=\pm~2

The set of solutions of that satisfies both affirmations is:

S=\{(x,~y)\in\mathbb{R}^2~|~(x,~y)=(-2,\,-2),~(2,~2)\}

This is the set we were looking for.

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