What two functions does the brain control and when we sleep

Mathematics

1 answer:

DENIUS [597]3 years ago

8 0

Answer:

circadian rhythm and homeostasis–work together to regulate when you are awake and sleep

Step-by-step explanation:

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Help on this problem?? please?

Reptile [31]

MrBillDoesMath!

Answer to #4: 81/256 * s^8 * t^ 12

Comments:

(7x^3) ^ (1/2) = 7 ^ (1/2) * x^(3/2) where ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.

---------------------

(1) (27s^7t^11)^ (4/3)

= 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)

As 27 = 3^3, 27 ^(4/3) = 3^4 = 81

(2) (-64st^2)^ (4/3) = (-64)^(4/3) * (s^4/3) * t(^8/3)

As 64 = (-4)^3, (-64)^(4/3) = (-4)^4 = +256

So (1)/(2) =

81 * s^(28/3)* t^(44/3)

------------------------------- =

256 s^(4/3) * t^((8/3)

81/256 * s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =

81/256 * s^(24/3) * t (36/3) =

81/256 * s^8 * t^ 12

MrB

5 0

3 years ago

What’s the domain and range of:<br> log(√(2x-1) + 3 )<br> Please explain how you got it too!!

Radda [10]

Two main facts are needed here:

1. The logarithm $\log x$ , regardless of the base of the logarithm, exists for $x>0$ .

2. The square root $\sqrt x$ exists for $x\ge0$ .

(in both cases we're assuming real-valued functions only)

By (2) we know that $\sqrt{2x-1}$ exists if $2x-1\ge0$ , or $x\ge\dfrac12$ .

By (1), we know that $\log(\sqrt{2x-1}+3)$ exists if $\sqrt{2x-1}+3>0$ , or $\sqrt{2x-1}>-3$ . But as long as the square root exists, it will always be positive, so this condition will always be met.

Ultimately, then, we only require $x\ge\dfrac12$ , so the function has domain $\left[\dfrac12,\infty)$ .

To determine the range, we need to know that, in their respective domains, $\sqrt x$ and $\log x$ increase monotonically without bound. We also know that $x=\dfrac12$ at minimum, at which point the square root term vanishes, so the least value the function takes on is $\log3$ . Then its range would be $[\log3,\infty)$ .