Step 1; First, we must convert all the temperatures into a common scale. The given temperatures are in three different units i.e. °C, K, and °F. Here we convert all the units into K.

To convert °C to K, we add 273.15 with the °C , 44 °C= 44 + 273.15 = 317.15 K.

To convert °F, we have the formula (112°F − 32) × ( $\frac{5}{9}$ ) + 273.15, so

110 °F = (110 - 32) × ( $\frac{5}{9}$ ) + 273.15 = 316.483 K,

112 °F = (112 - 32) × ( $\frac{5}{9}$ ) + 273.15 = 317.594 K.

Step 2; By rewriting all the values on the same unit, we get;

Sample A = 317.15 K,

Sample B = 320 K,

Sample C = 316.483 K,

Sample D = 310 K,

Sample E = 317.594.

The descending order is as follows 320 K, 317.594K, 317.15 K, 316.483 K and 310 K which is Sample B, Sample E, Sample A, Sample C, and Sample D.

5 0

4 years ago

What is the value of (fog) (-5)

DochEvi [55]

$f(x)=2x+1,\ g(x)=x^2\\\\(f\circ g)(x)=2(x^2)+1=2x^2+1\\\\(f\circ g)(-5)=2(-5)^2+1=2(25)+1=50+1=51$

6 0

4 years ago

How do you solve this?

k0ka [10]

$x^2-3x-4=0\\\\x^2-2x\cdot\frac{3}{2}=4\ \ \ /+(\frac{3}{2})^2\\\\x^2-2x\cdot\frac{3}{2}+(\frac{3}{2})^2=4+(\frac{3}{2})^2\\\\(x-\frac{3}{2})^2=4+\frac{9}{4}\\\\(x-\frac{3}{2})^2=\frac{16}{4}+\frac{9}{4}\\\\(x-\frac{3}{2})^2=\frac{25}{4}\\\\x-\frac{3}{2}=\pm\sqrt\frac{25}{4}$

$x-\frac{3}{2}=\pm\frac{\sqrt{25}}{\sqrt4}\\\\x-\frac{3}{2}=\pm\frac{5}{2}\\\\x-\frac{3}{2}=-\frac{5}{2}\ or\ x-\frac{3}{2}=\frac{5}{2}\\\\x=-\frac{5}{2}+\frac{3}{2}\ or\ x=\frac{5}{2}+\frac{3}{2}\\\\x=-\frac{2}{2}\ or\ x=\frac{8}{2}\\\\x=-1\ or\ x=4$

5 0

3 years ago

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