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Oduvanchick [21]
4 years ago
10

What is the domain of g? I need Very quickly please

Mathematics
1 answer:
Blizzard [7]4 years ago
5 0

Answer:

Domain: [-6, 6]

Step-by-step explanation:

Domain is the set of x-values that can be inputted into function <em>g</em>.

We see from the graph that our x-values span from -6 to 6. We also see that both points are also closed dots, so we use brackets to denote that they are included. Therefore, [-6, 6] would be our domain.

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Answer: 26 years

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3 years ago
Can anybody help me on this
daser333 [38]

Answer:

(A) 120°

Step-by-step explanation:

m∠P=x

m∠S=2x

m∠Q=x

m∠R=2x

Sum of the angles in trapezoid  is 360°.

x+x+2x+2x=360

6x=360

x=60°

m∠R=2x=2*60=120

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3 years ago
Betty has several of the standard six sided dice that are common in many board games if Betty rolls one of these dice what is th
AlexFokin [52]
1/6, because since there are six sides, and you need one of them to land, it is a 1/6 probability. 
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4 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
A teacher interested in determining the effect of a new computer program on learning to read conducted a study. One hundred stud
anzhelika [568]
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