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3 years ago

Absolute zero is shown as 0 on which scale?

Mathematics

2 answers:

Angelina_Jolie [31]3 years ago

4 0

In The Kelvin Scale
Hope it helps

mr_godi [17]3 years ago

3 0

Absolute zero is shown as 0 on the Kelvin Scale.

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How do you prove cosx = 1-tan^2(x/2)/1+tan^2(x/2)?

nikdorinn [45]

$\bf tan\left(\cfrac{{{ \theta}}}{2}\right)=
\begin{cases}
\pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}}
\\ \quad \\
\cfrac{1-cos({{ \theta}})}{sin({{ \theta}})}
\end{cases}\\\\
-------------------------------\\\\
tan^2\left( \frac{x}{2} \right)\implies \left[ \cfrac{sin(x)}{1+cos(x)} \right]^2\implies \cfrac{sin^2(x)}{[1+cos(x)]^2}
\\\\\\
\boxed{\cfrac{sin^2(x)}{1+2cos(x)+cos^2(x)}}$

now, let's plug that in the right-hand-side expression,

$\bf cos(x)=\cfrac{1-tan^2\left( \frac{x}{2} \right)}{1+tan^2\left( \frac{x}{2} \right)}\\\\
-------------------------------\\\\
\cfrac{1-tan^2\left( \frac{x}{2} \right)}{1+tan^2\left( \frac{x}{2} \right)}\implies \cfrac{1-\frac{sin^2(x)}{1+2cos(x)+cos^2(x)}}{1+\frac{sin^2(x)}{1+2cos(x)+cos^2(x)}}
\\\\\\
\cfrac{\frac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)}}{\frac{1+2cos(x)+cos^2(x)~+~sin^2(x)}{1+2cos(x)+cos^2(x)}}$

$\bf \cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{\underline{1+2cos(x)+cos^2(x)}}\cdot \cfrac{\underline{1+2cos(x)+cos^2(x)}}{1+2cos(x)+cos^2(x)~+~sin^2(x)}
\\\\\\
\cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)~+~sin^2(x)}$

$\bf -------------------------------\\\\
recall\qquad sin^2(\theta)+cos^2(\theta)=1\\\\
-------------------------------\\\\
\cfrac{\boxed{sin^2(x)+cos^2(x)}+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+\boxed{1}}
\\\\\\
\cfrac{cos^2(x)+2cos(x)+cos^2(x)}{2+2cos(x)}\implies \cfrac{2cos(x)+2cos^2(x)}{2+2cos(x)}
\\\\\\
\cfrac{\underline{2} cos(x)~\underline{[1+cos(x)]}}{\underline{2}~\underline{[1+cos(x)]}}\implies cos(x)$

8 0

3 years ago

Colby graphed a scatter plot of student exam scores (y) and the number of hours each student had slept the night before the exam

Nat2105 [25]

Answer: 612

Step-by-step explanation:

Substitute the value 6 (hours) in the equation and solve.

Y = 92x+60

Y = (92*6)+60

Y=552+60 = 612

8 0

3 years ago

Read 2 more answers

What value of b will cause the system to have an infinite

kvasek [131]

Hi there!

The way that an infinite number of solutions is achieved is when the two equations, when solved together, get an answer which is just a number equal to the same number. To do this, first substitute 6x-b in for y in the second equation.

$-3x+\frac{1}{2}(6x-b)=-3$

$-3x+3x-\frac{1}{2}b=-3$

$-\frac{1}{2}b=-3$

Now, we see that b needs to be equal to -3 when multiplied by -1/2. When both sides are divided by -1/2, b becomes equal to 6. Thus, b must be 6.

Check work:

$6x-6=y$

$-3x+\frac{1}{2}(6x-6)=-3$

$-3x+3x-3=-3$

$-3=-3$

Thus, as they are equal to each other, the answer is correct. b = 6.

Hope this helps!

8 0

4 years ago

I need to find the zeros to these two problems and I have no idea how

Iteru [2.4K]

If you have a graphing calculator just put in the equation in 'y=' (not the i equation), and then go to 2nd trace and see where the y=0, those numbers under the x column are the zeros. For the first one, the zeros are: -1, .5, and 2.8. For the second question the zeros are: -3 and about 1.9. The zeros with a decimal are estimations.

7 0

3 years ago

What are the vertex, axis of symmetry, domain and range of the function:

DiKsa [7]

Y=(-x*x)-4x+3
y=-(x^2+4x-3)
4+y=-(x^2+4x+4)+3
y+4=-(x-2)^2.) +3
-4. -4
y=-((x-2)^2)-1

Answers:
Vertex: (2,-1)
AOS:x = 2
Domain: All Real Numbers
Range:[-1,Infinity)

3 0

3 years ago