Answer: 2nd answer
Step-by-step explanation:
Kevin should have used a 4 in the numerator because there are 2 red sections and 2 green sections.
-4
The elephant lost 4 pounds each month. Because he is losing weight, and not gaining weight, you have to make the answer negative.
Answer:
AC = XY
Step-by-step explanation:
A triangle is a polygon with three sides and three angles. The types of triangles are scalene triangle, equilateral triangle, right angled triangle and obtuse triangle.
Two triangles are said to be congruent if all the three sides and three angles of one triangle is equal to the three sides and three angles of the other triangle.
In Δ ABC and ΔXYZ ,∠B=∠X=90. This means that triangle ABC and triangle XYZ are right angled triangle since one of their angles is 90 degrees.
The hypotenuse is the side opposite to right angle (90 degrees).
Therefore AC is the hypotenuse of triangle ABC and YZ is the hypotenuse of triangle XYZ.
If the hypotenuse and one leg of a right angled triangle is equal to the hypotenuse and leg of another right angled triangle, then the two triangles are congruent by RHS rule.
Since BC = XY, for Δ ABC congruent to ΔXYZ by R.H.S condition, the hypotenuse have to be equal, i.e. AC = XY
Answer:
ani pong kaseng tanong to ang gulo hindi ko po maintindihan
Answer and explanation:
Given : Consider the Ideal Gas Law,
where k>0 is a constant.
To find : Solve this equation for V in terms of P and T.
Solution :
![PV=kT](https://tex.z-dn.net/?f=PV%3DkT)
Divide each side by P,
....(1)
a) Determine the rate of change of the volume with respect to the pressure at constant temperature. Interpret the result.
Differentiate equation (1) w.r.t P,
![\frac{dV}{dP}=kT\frac{d}{dP}(\frac{1}{P})](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BdP%7D%3DkT%5Cfrac%7Bd%7D%7BdP%7D%28%5Cfrac%7B1%7D%7BP%7D%29)
![\frac{dV}{dP}=kT(-\frac{1}{P^2})](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BdP%7D%3DkT%28-%5Cfrac%7B1%7D%7BP%5E2%7D%29)
![\frac{dV}{dP}=-\frac{kT}{P^2}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BdP%7D%3D-%5Cfrac%7BkT%7D%7BP%5E2%7D)
b) Determine the rate of change of the volume with respect to the temperature at constant pressure. Interpret the result.
Differentiate equation (1) w.r.t T,
![\frac{dV}{dT}=\frac{k}{P}\frac{d}{dT}(T)](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BdT%7D%3D%5Cfrac%7Bk%7D%7BP%7D%5Cfrac%7Bd%7D%7BdT%7D%28T%29)
![\frac{dV}{dP}=\frac{k}{P}(1)](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BdP%7D%3D%5Cfrac%7Bk%7D%7BP%7D%281%29)
![\frac{dV}{dP}=\frac{k}{P}](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7BdP%7D%3D%5Cfrac%7Bk%7D%7BP%7D)
c) Assuming k =1, draw several level curves of the volume function and interpret the results.
When k=1, ![PV=T](https://tex.z-dn.net/?f=PV%3DT)
<0
>0
Refer the attached figure below.