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11 months ago

12-6r=2r+36 (2-step equations)

Mathematics

2 answers:

mart [117]11 months ago

5 0

Answer: r=-3

Step-by-step explanation:

12-6r=2r+36

12=8r+36

-24=8r

r=-3

ad-work [718]11 months ago

3 0

The answer is -3.

Explation:

12-6r=2r+36
+6r +6r
———————
12=8r+36
-36 -36

-24=8r
then divide each side by 8 so you then get r= -3 as you’re answer.

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Use the distributive property to write an equivalent expression in simplest form. 9(4h+ 9m)

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A deli made $320,000 in revenue last year. Due to increased popularity of the deli, the owner predicts that revenue will increas

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Step-by-step explanation:

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2 years ago

A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t

11Alexandr11 [23.1K]

Answer:

Question 1: $dm/dt=0.2m$

Question 2: $m=Ae^{(0.2t)}$

Question 3: $m=10e^{(0.2t)}$

Question 4: $m=10g$

Explanation:

Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

Question 2: Find the general solution of this equation. Use A as a constant of integration.

a) Separate variables

$dm/m=0.2dt$

b) Integrate

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) Antilogarithm

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

Use the measured rate of 4 grams per hour after 3 hours

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First, find the mass at t = 3 hours

$dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g$

Now substitute in the general solution of the differential equation, to find A:

$m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976$

Round A to 1 significant figure:

A = 10.

Particular solution:

$m=10e^{(0.2t)}$

Question 4. What was the mass of the bacteria at time =0?

Substitute t = 0 in the equation of the particular solution:

$m=10e^{0}\\\\m=10g$

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2 years ago

A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th

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Answer:

The rocket hits the ground at a time of 11.59 seconds.

Step-by-step explanation:

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It hits the ground when $y = 0$ , so we have to find x for which $y = 0$ , which is a quadratic equation.

Finding the roots of a quadratic equation:

Given a second order polynomial expressed by the following equation:

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This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

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