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3 years ago

Is a 39 Celsius body temperature higher than normal 98 Fahrenheit?? Yes or No

Mathematics

1 answer:

Umnica [9.8K]3 years ago

5 0

Answer:

Yes.

Step-by-step explanation:

According to medical science, a body temperature considered normal (that is, below fever) in degrees Fahrenheit is 98ºF, while its equivalent in degrees Celsius is 36.6ºC. Therefore, a body temperature of 39ºC is higher than 98ºF, being equivalent to 102.2ºF.

This is verified by applying the conversion formula from Celcius to Fahrenheit:

(39 x 9/5) + 32 = X

39 x 1.8 + 32 = X

70.2 + 32 = X

102.2 = X

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Morgarella [4.7K]

Answer:

$f'(x)=-\frac{2}{x^\frac{3}{2}}$

Step-by-step explanation:

So we have the function:

$f(x)=\frac{4}{\sqrt x}$

And we want to find the derivative using the limit process.

The definition of a derivative as a limit is:

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Therefore, our derivative would be:

$\lim_{h \to 0}\frac{\frac{4}{\sqrt{x+h}}-\frac{4}{\sqrt x}}{h}$

First of all, let's factor out a 4 from the numerator and place it in front of our limit:

$=\lim_{h \to 0}\frac{4(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x})}{h}$

Place the 4 in front:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}$

Now, let's multiply everything by (√(x+h)(√(x))) to get rid of the fractions in the denominator. Therefore:

$=4\lim_{h \to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}(\frac{\sqrt{x+h}\sqrt x}{\sqrt{x+h}\sqrt x})$

Distribute:

$=4\lim_{h \to 0}\frac{({\sqrt{x+h}\sqrt x})\frac{1}{\sqrt{x+h}}-(\sqrt{x+h}\sqrt x)\frac{1}{\sqrt x}}{h({\sqrt{x+h}\sqrt x})}$

Simplify: For the first term on the left, the √(x+h) cancels. For the term on the right, the (√(x)) cancel. Thus:

$=4 \lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }$

Now, multiply both sides by the conjugate of the numerator. In other words, multiply by (√x + √(x+h)). Thus:

$= 4\lim_{h\to 0}\frac{\sqrt x-(\sqrt{x+h})}{h(\sqrt{x+h}\sqrt{x}) }(\frac{\sqrt x +\sqrt{x+h})}{\sqrt x +\sqrt{x+h})}$

The numerator will use the difference of two squares. Thus:

$=4 \lim_{h \to 0} \frac{x-(x+h)}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Simplify the numerator:

$=4 \lim_{h \to 0} \frac{x-x-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}\\=4 \lim_{h \to 0} \frac{-h}{h(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Both the numerator and denominator have a h. Cancel them:

$=4 \lim_{h \to 0} \frac{-1}{(\sqrt{x+h}\sqrt x)(\sqrt x+\sqrt{x+h})}$

Now, substitute 0 for h. So:

$=4 ( \frac{-1}{(\sqrt{x+0}\sqrt x)(\sqrt x+\sqrt{x+0})})$

Simplify:

$=4( \frac{-1}{(\sqrt{x}\sqrt x)(\sqrt x+\sqrt{x})})$

(√x)(√x) is just x. (√x)+(√x) is just 2(√x). Therefore:

$=4( \frac{-1}{(x)(2\sqrt{x})})$

Multiply across:

$= \frac{-4}{(2x\sqrt{x})}$

Reduce. Change √x to x^(1/2). So:

$=-\frac{2}{x(x^{\frac{1}{2}})}$

Add the exponents:

$=-\frac{2}{x^\frac{3}{2}}$

And we're done!

$f(x)=\frac{4}{\sqrt x}\\f'(x)=-\frac{2}{x^\frac{3}{2}}$

5 0

3 years ago

The discount on a toy is 40% the toy robot is on sale for $54 find the original price

maria [59]

Answer:

The original price is $90.

Step-by-step explanation:

100% - 40% = 60%

$\frac{54}{y} :\frac{60}{100}$

y × 60 = 54 × 100

60y = 5400

60y ÷ 60 = 5400 ÷ 60

y = 90

5 0

3 years ago

Read 2 more answers

In the school library, 28% of the books are non-fiction. If there are 1,224 fiction books in the library, what is the total numb

N76 [4]

Answer:

<em>There are 1700 books in the library.</em>

Step-by-step explanation:

<em>If 28% of the books is non-fiction, that means the remaining 72% of books are fiction.</em>

If 72% of the books are fiction, 1224/72 will give us 1%, which is 17.

If 1% is 17 books, then 17*100=100% of books.

There are 1700 books in the library.

8 0

3 years ago

Which of the following Could be used to determine the volume of a sheet of metal 50×34 that had identical squares cut out of the

valentinak56 [21]

Answer:

idk

Step-by-step explanation:

8 0

3 years ago

How many possible outcomes are there when flipping a coin nine times

Vedmedyk [2.9K]

18 is my best answer

6 0

3 years ago