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3 years ago

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The temperature of a cooling liquid over time can be modeled by the exponential function T(x) = 60(1/2)^(x/30) + 20, where T(x)

is the temperature in degrees Celsius, and x is the elapsed time in minutes. Graph the function and determine how long it takes for the temperature to reach 28°C. What was the initial temperature?

1 answer:

Alexus [3.1K]3 years ago

6 0

Based on the scenario, the initial temperature would be :

28 = (60) (1/2)^x/30 + 20

(60) (1/2)^0/30 + 20 = 60 + 20 =

80 C

Hope this helps

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Give a geometric description of the following system of equations.a. 2x−4y=12 −3x+6y=−15.b. 2x−4y=12 −5x+3y=10.a. 2x−4y=12 −3x+6

Reptile [31]

Answer:

a. No solution, parallel lines.

b. One solution.

Step-by-step explanation:

Given the system of equations:

a. $2x-4y=12$

$-3x+6y=-15$

b. $2x-4y=12$

$-5x+3y=10$

To give a geometric description of the given system of equations.

The geometric description of a system of equations in 2 variables mean the system of equations will represent the number of lines equal to the number of equations in the system given.

i.e.

Number of planes = Number of variables

Number of lines = Number of equations in the system.

Here, we are given 2 variables and 2 equation in each system.

So, they can be represented in the xy-coordinates plane.

And the number of solutions to the system depends on the following condition.

Let the system of equations be:

$A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0$

1. One solution:

There will be one solution to the system of equations, If we have:

$\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}$

2. Infinitely Many Solutions: (Identical lines in the system)

$\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}= \dfrac{C_1}{C_2}$

3. No Solution:(Parallel lines)

$\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}$

Now, let us discuss the system of equations one by one:

a. $2x-4y=12$ OR $2x-4y-12=0$

$-3x+6y=-15$ OR $-3x+6y+15=0$

$A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -3, B_2 = 6, C_2= 15$

Here, the ratio:

$\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2} = -\dfrac{2}{3}\\\dfrac{C_1}{C_2} = -\dfrac{4}{5}$

$\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}$

Therefore, no solution i.e. parallel lines.

b. $2x-4y=12$ OR $2x-4y-12=0$

$-5x+3y=10$ OR $-5x+3y-10=0$

$A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -5, B_2 = 3, C_2 = -10$

$\dfrac{A_1}{A_2}= -\dfrac{2}{5}\\\dfrac{B_1}{B_2} = -\dfrac{4}{3}\\\dfrac{C_1}{C_2} = -\dfrac{6}{5}$

$\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}$

So, one solution.

Kindly refer to the images attached for the graphical representation of the given system of equations.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx%20%5Csqrt%7Bx....%7D%20%7D%20%7D%20%7D%20%20%3D%20%

andrey2020 [161]

First observe that if $a+b>0$ ,

$(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}$

Let $a=0$ and $b=x$ . It follows that

$a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}$

Now let $b=1$ , so $a^2+a=4x$ . Solving for $a$ ,

$a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2$

which means

$a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}$

Now solve for $x$ .

$x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2$

(note that we assume $2x-1\ge0$ )

$4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}$

(we omit $x=0$ since $2\cdot0-1=-1\ge0$ is not true)

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1 year ago

The amount of snowfall over 6 months in Michigan was 2, 12, 8, 18, 8, and 0 inches. find the range and median of the monthly sno

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Step-by-step explanation:

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