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gtnhenbr [62]
2 years ago
8

Need it solved correctly for khan academy

Mathematics
1 answer:
RUDIKE [14]2 years ago
6 0

The tiger population loses 3/5 of its size every 2.94 decades

<h3>Rate of change using differential calculus</h3>

The given equation is:

N(t)=710(\frac{8}{125} )^t

Find the derivative of the given function

\frac{dN}{dt} =710(0.064)^tln(0.064)\\\\\frac{dN}{dt} =-1951.7(0.064)^t

When the tiger loses 3/5 of its population

dN/dt = 3/5

Solve for t

\frac{3}{5} =-1951.7(0.064)^t\\\\-0.0003=(0.064)^t

Take the natural logarithm of both sides

ln(-0.0003)=t(ln0.064)\\\\-8.087=-2.75t\\\\t=\frac{-8.087}{-2.75} \\\\t=2.94

The tiger population loses 3/5 of its size every 2.94 decades

Learn more on rate of change using calculus here: brainly.com/question/96116

#SPJ1

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Match the hyperbolas represented by the equations to their foci.
Arte-miy333 [17]

Answer:

1) (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- the coordinates of the foci are (h ± c , k), where c² = a² + b²

- The standard form of the equation of a hyperbola with

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

- the coordinates of the foci are (h , k ± c), where c² = a² + b²

* Lets look to the problem

1) The foci are (1 , -22) and (1 , 12)

- Compare the point with the previous rules

∵ h = 1 and k ± c = -22 ,12

∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1

∵ k + c = -22 ⇒ (1)

∵ k - c = 12 ⇒ (2)

* Add (1) and(2)

∴ 2k = -10 ⇒ ÷2

∴ k = -5

* substitute the value of k in (1)

∴ -5 + c = -22 ⇒ add 5 to both sides

∴ c = -17

∴ c² = (-17)² = 289

∵ c² = a² + b²

∴ a² + b² = 289

* Now lets check which answer has (h , k) = (1 , -5)

  and a² + b² = 289 in the form (y - k)²/a² - (x - h)²/b² = 1

∵ 15² + 8² = 289

∵ (h , k) = (1 , -5)

∴ The answer is (y + 5)²/15² - (x - 1)²/8² = 1

* (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) The foci are (-7 , 5) and (3 , 5)

- Compare the point with the previous rules

∵ k = 5 and h ± c = -7 ,3

∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1

∵ h + c = -7 ⇒ (1)

∵ h - c = 3 ⇒ (2)

* Add (1) and(2)

∴ 2h = -4 ⇒ ÷2

∴ h = -2

* substitute the value of h in (1)

∴ -2 + c = -7 ⇒ add 2 to both sides

∴ c = -5

∴ c² = (-5)² = 25

∵ c² = a² + b²

∴ a² + b² = 25

* Now lets check which answer has (h , k) = (-2 , 5)

  and a² + b² = 25 in the form (x - h)²/a² - (y - k)²/b² = 1

∵ 3² + 4² = 25

∵ (h , k) = (-2 , 5)

∴ The answer is (x + 2)²/3² - (y - 5)²/4² = 1

* (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) The foci are (-6 , -2) and (14 , -2)

- Compare the point with the previous rules

∵ k = -2 and h ± c = -6 ,14

∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1

∵ h + c = -6 ⇒ (1)

∵ h - c = 14 ⇒ (2)

* Add (1) and(2)

∴ 2h = 8 ⇒ ÷2

∴ h = 4

* substitute the value of h in (1)

∴ 4 + c = -6 ⇒ subtract 4 from both sides

∴ c = -10

∴ c² = (-10)² = 100

∵ c² = a² + b²

∴ a² + b² = 100

* Now lets check which answer has (h , k) = (4 , -2)

  and a² + b² = 100 in the form (x - h)²/a² - (y - k)²/b² = 1

∵ 8² + 6² = 100

∵ (h , k) = (4 , -2)

∴ The answer is (x - 4)²/8² - (y + 2)²/6² = 1

* (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) The foci are (-7 , -10) and (-7 , 16)

- Compare the point with the previous rules

∵ h = -7 and k ± c = -10 , 16

∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1

∵ k + c = -10 ⇒ (1)

∵ k - c = 16 ⇒ (2)

* Add (1) and(2)

∴ 2k = 6 ⇒ ÷2

∴ k = 3

* substitute the value of k in (1)

∴ 3 + c = -10 ⇒ subtract 3 from both sides

∴ c = -13

∴ c² = (-13)² = 169

∵ c² = a² + b²

∴ a² + b² = 169

* Now lets check which answer has (h , k) = (-7 , 3)

  and a² + b² = 169 in the form (y - k)²/a² - (x - h)²/b² = 1

∵ 5² + 12² = 169

∵ (h , k) = (-7 , 3)

∴ The answer is (y - 3)²/5² - (x + 7)²/12² = 1

* (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

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3 years ago
Want Money. Answer thisss below.
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weqwewe [10]

2x+6 - х -4

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. Benson had -$12 in his bank
natka813 [3]

Answer: 24

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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The two cases together weigh 102,100 decigrams.

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