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DaniilM [7]
2 years ago
14

Write an equivalent expression in expanded form. 8(-x+1/4)

Mathematics
1 answer:
professor190 [17]2 years ago
7 0

Answer:

-8x + 2

Step-by-step explanation:

8(-x+1/4)

(8*(-x)] +(8 x 1/4)

-8x + 2

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Answer:

  [-4,3]

Step-by-step explanation:

The domain is the left-right (horizontal) extent of the function. The dots at the end of the continuous line are solid, so the brackets on the interval are square brackets.

The left dot is at x=-4; the right dot is at x=3, so the domain is ...

  [-4,3]

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Step-by-step explanation:

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Write an equation that represents the following statement: 6 less than the product of r and 4 is 2
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Answer: R x 4 - 6

Step-by-step explanation:

8 0
3 years ago
A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and
allsm [11]

(a) Let A(t) denote the amount of sugar in the tank at time t. The tank starts with only pure water, so \boxed{A(0)=0}.

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(<em>A(t)</em>/1080 kg/L) * (7 L/min) = 7<em>A(t)</em>/1080 kg/min

so that the net rate of change of A(t) is governed by the ODE,

\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}

or

A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}

Multiply both sides by the integrating factor e^{7t/1080} to condense the left side into the derivative of a product:

e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}

\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}

Integrate both sides:

e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt

e^{\frac{7t}{1080}}A(t)=\dfrac{378}5e^{\frac{7t}{1080}}+C

Solve for A(t):

A(t)=\dfrac{378}5+Ce^{-\frac{7t}{1080}}

Given that A(0)=0, we find

0=\dfrac{378}5+C\implies C=-\dfrac{378}5

so that the amount of sugar at any time t is

\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}

(c) As t\to\infty, the exponential term converges to 0 and we're left with

\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5

or 75.6 kg of sugar.

7 0
3 years ago
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