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4 years ago

Analyze the linear inequalities and determine if the solution set is the shaded region above or below the boundary line.

Mathematics

1 answer:

KengaRu [80]4 years ago

7 0

Hello!

If the y side is greater than the x side then it is shaded above

If the x side is greater than the y side then it is shaded below

So the answers are

above, above, below, above, below, below

Hope this helps!

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The Miami airport offers several different parking options for visitors. A new short term

Lana71 [14]

Answer:

(0, 24]

Step-by-step explanation:

The horizontal extent of the graph is from 0 to 24. There is an open circle at x=0, and a closed circle at x=24. Hence, the domain of the function is ...

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In interval notation, this is

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4 years ago

Find the solution to the differential equation dz/dt = 7te^4z that passes through the origin.

andreev551 [17]

Answer:

$z=(\frac{-1}{4} )ln|1-14t^{2}|\\$

Step-by-step explanation:

from the equation, $\frac{dz}{dt}=7te^{4z}\\$ .

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$\frac{dz}{e^{4z}}=7tdt\\e^{-4z} dz=7tdt$ .

if we integrate both side,

$\int\limits^a_b{e^{-4z} } \,dz =\int\limits^a_b {7t} \,dt$

$-1/4e^{-4z} +c_{1}=7/2t^{2} +c_{2}\\-1/4e^{-4z}= 7/2t^{2} +c_{2}-c_{1}\\let c_{2}-c_{1}=c \\-1/4e^{-4z}= 7/2t^{2} +c$

since the equation passes through the origin, and at the origin z=0 and t=0

we substitute this values and solve for the constant c

$e^{-4*0}= 7/2*0 +c\\c=1$ .

If we substitute the value of c into the equation we arrive at

$(-1/4)e^{-4z}= (7/2)t^{2}+1\\ e^{-4z}=1-14t^{2} \\$

if we the the natural logarithm of both sides, we arrive at

$-4z=ln|1-14t^{2}|\\z=(\frac{-1}{4} )ln|1-14t^{2}|\\$

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3 years ago

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3 years ago

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Rama09 [41]

Answer:

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3 years ago

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Ilia_Sergeevich [38]

Answer:

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Step-by-step explanation:

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