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2 years ago

An element with mass 310 grams decays by 8. 9% per minute. How much of the element is remaining after 19 minutes, to the nearest

10th of a gram?.

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

3 0

$\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\\ P=\textit{initial amount}\dotfill &310\\ r=rate\to 8.9\%\to \frac{8.9}{100}\dotfill &0.089\\ t=\textit{elapsed time}\dotfill &19\\ \end{cases} \\\\\\ A=310(1-0.089)^{19}\implies A=310(0.911)^{19}\implies A\approx 52.8$

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DanielleElmas [232]

Answer:

Step-by-step explanation:

Volume can be described as the total amount of space that is in a closed surface. this amount of space is usually three dimensional. Volume is usually measured in m³ or cm³ or in³

For example, the volume of a rectangular box given the following dimension : length - 3cm

breadth - 8cm

height - 10 cm

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3 years ago

K is the midpoint of PQ, P has

Pachacha [2.7K]

Answer:

Coordinates of Q $(x_2,y_2) \:are\: \mathbf{(7,16)}$

Option D is correct option.

Step-by-step explanation:

We are given:

K is the midpoint of PQ

Coordinates of P = (-9,-4)

Coordinates of K = (-1,6)

We need to find coordinates of Q $(x_2,y_2)$

We will use the formula of midpoint: $Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We are given midpoint K and $x_1,y_1$ the coordinates of P we need to find $x_2,y_2$ the coordinates of Q.

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\(-1,6)=(\frac{-9+x_2}{2},\frac{-4+y_2}{2})\\$

Now, we can write

$-1=\frac{-9+x_2}{2}, 6=\frac{-4+y_2}{2}\\Simplifying:\\-2=-9+x_2\:,\: 12=-4+y_2\\-2+9=x_2\:,\: 12+4=+y_2\\x_2=7\:,\:y_2=16$

So, we get coordinates of Q $(x_2,y_2) \:are\: \mathbf{(7,16)}$

Option D is correct option.

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3 years ago

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3 years ago

A box with a hinged lid is to be made out of a rectangular piece of cardboard that measures 3 centimeters by 5 centimeters. Six

kherson [118]

Answer:

x = 0.53 cm

Maximum volume = 1.75 cm³

Step-by-step explanation:

Refer to the attached diagram:

The volume of the box is given by

$V = Length \times Width \times Height \\\\$

Let x denote the length of the sides of the square as shown in the diagram.

The width of the shaded region is given by

$Width = 3 - 2x \\\\$

The length of the shaded region is given by

$Length = \frac{1}{2} (5 - 3x) \\\\$

So, the volume of the box becomes,

$V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\$

In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.

$\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\$

We are left with a quadratic equation.

We may solve the quadratic equation using quadratic formula.

The quadratic formula is given by

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where

$a = 18 \\\\b = -38 \\\\c = 15 \\\\$

$x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\$