35.) The temperature of dry ice is -109.3°F.

Mathematics

1 answer:

kari74 [83]2 years ago

4 0

Hello there!

You're given the value of -109.3°F, and you are told that is 184.9°F less than the outside temperature. To find the outside temperature, add the difference to the temperature of dry ice.

-109.3°F + 184.9°F =75.6°F

So, the outside temperature is 75.6°F. I hope this was helpful, have a great rest of your day!

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Determine whether the given function is a solution to the given differential equation. y=cosx+x8, d^2y/dx^2 + y = x^8 + 56x^6.

Ulleksa [173]

The correct question is:

Determine whether the given function is a solution to the given differential equation. y = cosx + x^8; d²y/dx² + y = x^8 + 56x^6

Step-by-step explanation:

Given the differential equation

d²y/dx² + y = x^8 + 56x^6.

Suppose y = cosx + x^8 is a solution, then differentiating y twice, and adding it to itself, must give the value on the right hand side of the differential equation.

Let us differentiate y twice

y = cosx + x^8

dy/dx = -sinx + 8x^7

d²y/dx² = -cosx + 56x^6

Now,

d²y/dx² + y = -cosx + 56x^6 + cosx + x^8

= 56x^6 + x^8

Therefore,

d²y/dx² + y = x^8 + 56x^6

Which shows that y = cosx + x^8 is a solution to the differential equation.

4 0

2 years ago