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andrew-mc [135]
3 years ago
12

The aquarium has 6 fewer yellow fish than green fish. 40 percent of the fish are yellow. How many green fish are in the aquarium

? Show your work.
Mathematics
1 answer:
Brilliant_brown [7]3 years ago
7 0
Y = number of yellow fish
G = number of green fish

Y = G - 6
Y = 0.40 * (Y+G)

G - 6 = 0.4*(G-6+G)
G - 6 = 0.4*(2G-6)
G - 6 = 0.8G -2.4
0.2G = 3.6
G = 3.6/0.2=18


There are 18 green fish.
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larisa86 [58]

Since \cot(x)=\frac{2}{3} and \cot^{2} x+1=\csc^{2} x, we know that:

\left(\frac{2}{3} \right)^{2}+1=\csc^{2} x\\\\\frac{13}{9}=\csc^{2} x\\\\\csc x=\frac{\sqrt{13}}{3}

If \csc x=\frac{\sqrt{13}}{3}, this means that \sin x=\frac{3}{\sqrt{13}} and by the Pythagorean identity,

\sin^{2} x+\cos^{2} x=1\\\left(\frac{3}{\sqrt{13}} \right)^{2}+\cos^{2} x=1\\\frac{9}{13}+\cos^{2} x=1\\\cos^{2} x=\frac{4}13}\\\cos x=\frac{2}{\sqrt{13}}

  • Using the double angle formula for sine, \sin(2x)=2\left(\frac{3}{\sqrt{13}} \right)\left(\frac{2}{\sqrt{13}} \right)=\boxed{\frac{12}{13}}
  • Using the double angle formula for cosine, \cos(2x)=1-2\left(\frac{3}{\sqrt{13}} \right)^{2}=\boxed{-\frac{5}{13}}
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7 0
2 years ago
Jack's brother is 14 years older than him. In six years time, Jack's brother will be twice as old as him. How old is Jack now?
mr Goodwill [35]
14+14 is 28 28-6 is 22 so Jack is 22
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3 years ago
Find the approximate area of the regions bounded by the curves y = x/(√x2+ 1) and y = x^4−x. (You may use the points of intersec
Finger [1]

The approximate area of the region bounded by the curves f(x) = x / √(x² + 1) and g(x) = x⁴ - x is approximately 0.806.

<h3>How to determine the approximate area of the regions bounded by the curves</h3>

In this problem we must use definite integrals to determine the area of the region bounded by the curves. Based on all the information given by the graph attached below, the area can be defined in accordance with this formula:

A = A₁ + A₂                                                                (1)

A₁ = ∫ [g(x) - f(x)] dx, for x ∈ [- 0.786, 0]                   (2)

A₂ = ∫ [f(x) - g(x)] dx, for x ∈ [0, 1.151]                       (3)

g(x) = x⁴ - x                                                               (4)

f(x) = x / √(x² + 1)                                                      (5)

Then, we proceed to find the integrals:

∫ g(x) dx = ∫ x⁴ dx - ∫ x dx = (1 / 5) · x⁵ - (1 / 2) · x²                          (6)

∫ f(x) dx = ∫ [x / √(x² + 1)] dx = (1 / 2) ∫ [2 · x / √(x² + 1)] dx = (1 / 2) ∫ [du / √u] = √u = √(x² + 1)                                                                                  (7)

And the complete expression for the integral is:

A = A₁ + A₂                                                                                      (1b)

A₁ = (1 / 5) · x⁵ - (1 / 2) · x² - √(x² + 1), for x ∈ [- 0.786, 0]               (2b)

A₂ = √(x² + 1) - (1 / 5) · x⁵ + (1 / 2) · x², for x ∈ [0, 1.151]                  (3b)

A₁ = 0.023

A₂ = 0.783

A = 0.023 + 0.783

A = 0.806

The approximate area of the region bounded by the curves f(x) = x / √(x² + 1) and g(x) = x⁴ - x is approximately 0.806.

To learn more on definite integral: brainly.com/question/14279102

#SPJ1

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Hello there,

I hope you and your family are staying safe and healthy during this winter season.

x^2 + y^2 -6x+14y-1=0

We need to use the Quadratic Formula*

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Thus, given the problem:

a = 1, b=-6, c=y^2+14y-1

So now we just need to plug them in the Quadratic Formula*

x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2} , x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}

As you can see, it is a mess right now. Therefore, we need to simplify it

x=\frac{6+2\sqrt{10-y^2-14y} }{2}, x = \frac{6-2\sqrt{10-y^2-14y} }{2}

Now that's get us to the final solution:

x=3+\sqrt{10-y^2-14y}, x=3-\sqrt{10-y^2-14y}

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

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6 0
3 years ago
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