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AlladinOne [14]
3 years ago
9

On average, Tony the turtle completes a 15-foot course in 1 minute, and Antonio the turtle completes an 18-foot course in 1 minu

te. If Tony and Antonio both start at the same time and travel in the same direction, how far apart will the turtles be after 5 minutes?(1 point)
75 feet

15 feet

90 feet

165 feet
Mathematics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

15 feet.

Step-by-step explanation:

Tony's speed = 15 ft/min.

Antonio's = 18 feet /min.

So in 5 minutes Tony travels 5*15 = 75 feet and Antonio travels 5*18 = 90 feet.

So the answer is 90-75 = 15 feet.

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Step-by-step explanation:

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If x+y=12 and xy=15,find the value of (x^2+y^2)
Mice21 [21]

Answer:  x^2+y^2=\dfrac{105\pm 18\sqrt6}{2}

<u>Step-by-step explanation:</u>

EQ1:  x + y = 12     --> x = 12 - y

EQ2:  xy = 15      

Substitute x = 12-y into EQ2 to solve for y:

(12 - y)y = 15

12y - y² = 15

0 = y² - 12y + 15

    ↓     ↓         ↓

  a=1   b= -12   c=15

.\  y=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\.\quad =\dfrac{-(-12)\pm \sqrt{(-12)^2-4(1)(15)}}{2(1)}\\\\\\.\quad =\dfrac{12\pm \sqrt{144-120}}{2}\\\\\\.\quad =\dfrac{12\pm \sqrt{24}}{2}\\\\\\.\quad =\dfrac{12\pm 2\sqrt{6}}{2}\\\\\\.\quad =6\pm \sqrt{6}

Now, let's solve for x:

xy=15\\\\x(6\pm\sqrt6)=15\\\\x=\dfrac{15}{6\pm\sqrt6}\\\\\\x=\dfrac{15}{6\pm\sqrt6}\bigg(\dfrac{6\pm\sqrt6}{6\pm\sqrt6}\bigg)=\dfrac{6\pm \sqrt6}{2}

Lastly, find x² + y² :

y^2=(6\pm \sqrt6)^2\quad \rightarrow \quad y^2=36\pm 12\sqrt6 +6\quad \rightarrow \quad y^2=42\pm 12\sqrt6

x^2=\bigg(\dfrac{6\pm \sqrt6}{2}\bigg)^2\quad \rightarrow \quad x^2=\dfrac{42\pm 12\sqrt6}{4}\quad \rightarrow \quad x^2=\dfrac{21\pm 6\sqrt6}{2}

                                                                                 

x^2+y^2=\dfrac{21\pm 6\sqrt6}{2}+42\pm 12\sqrt6\\\\\\.\qquad \quad = \dfrac{21\pm 6\sqrt6}{2}+\dfrac{84\pm 24\sqrt6}{2}\\\\\\. \qquad \quad = \dfrac{105\pm 18\sqrt6}{2}

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Answer:

#3 a. g(-1) = 2, g(0) = 3, and g(1) = 2

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Step-by-step explanation:

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b. The given function g(x) = -x² + 3 for finding the value of <em>g</em> can take any value of <em>x</em> which is a real number

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h(x) = \dfrac{x}{x - 2}

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h(-1) = \dfrac{-1}{(-1) - 2} = \dfrac{-1}{-3} = \dfrac{1}{3}

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\therefore h(0) =0

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