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Afina-wow [57]
2 years ago
10

The pond can hold 400 water lilies. By what day will the pond be full? Write and solve an equation. The pond will be full by the

end of day.
Mathematics
2 answers:
egoroff_w [7]2 years ago
8 0

The pond will be full by the end of the day in 93 days.

<h3>Linear system</h3>

It is a system of an equation in which the highest power of the variable is always 1. A one-dimension figure that has no width. It is a combination of infinite points side by side.

Given

The pond can hold 400 water liters.

y = 3.915(1.106)x is an expression.

<h3>To find </h3>

The pond will be full by the end of the day.

<h3>How many days it will take?</h3>

Let x be the day and y be the hold of water.

When pond can hold 400 liters of water the x will be

400 = 3.915×1.106x

400 = 4.33 x

    x = 92.37 ≈ 93

Thus, the pond will be full by the end of the day in 93 days.

More about the linear system link is given below.

brainly.com/question/20379472

steposvetlana [31]2 years ago
4 0

Answer:

46

Step-by-step explanation:

just did it on edge

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Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi
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Answer:

b. circle; 2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0

Step-by-step explanation:

The given conic has equation;

x^2-5x+y^2=3

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This is a circle with center;

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This implies that;

x=\frac{5}{2},y=0

When the circle is rotated through an angle of \theta=\frac{\pi}{3},

The new center is obtained using;

x'=x\cos(\theta)+y\sin(\theta) and y'=-x\sin(\theta)+y\cos(\theta)

We plug in the given angle with x and y values to get;

x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3}) and y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})

This gives us;

x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}

The equation of the rotated circle is;

(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}

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Multiply through by 4; to get

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Write in general form;

4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0

Divide through by 2.

2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0

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