Set up two equations.
s + 12 = l , s + l = 74 (these equations represent the problem)
s + (s + 12) = 74 plug in "l" to the second equation.
2s + 12 = 74 combine like terms.
2s + 12 - 12 = 74 - 12 subtract to isolate the variable
2/2s = 62/2 divide to isolate the variable.
s = 31. This means the smaller number is 31.
31 + 12 = 43. since the larger number is 12 more than the smaller, add 12.
l = 43 the larger number has a value of 43.
small number = 31
larger number = 43
hope this helps! :D
Negative abscissa, zero ordinate is between Quadrant II and Quadrant III.
Hope this helps! :)
Answer:
384
Step-by-step explanation:
Answer:
The answer is c
Step-by-step explanation:
Well, since the Circle is 8cm you can divide the circle by two to get 4
![p= \frac{RT}{vm} + \frac{a+bT}{vm^{2}} \\or\\v= \frac{1}{p}[ \frac{R}{m} + \frac{a+bT}{m^{2}}]](https://tex.z-dn.net/?f=p%3D%20%5Cfrac%7BRT%7D%7Bvm%7D%20%2B%20%5Cfrac%7Ba%2BbT%7D%7Bvm%5E%7B2%7D%7D%20%5C%5Cor%5C%5Cv%3D%20%5Cfrac%7B1%7D%7Bp%7D%5B%20%5Cfrac%7BR%7D%7Bm%7D%20%20%2B%20%5Cfrac%7Ba%2BbT%7D%7Bm%5E%7B2%7D%7D%5D%20)
![\frac{\partial v}{\partial T} |_{p}= \frac{1}{p}[ \frac{R}{m}+ \frac{b}{m^{2}}]](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%20%7C_%7Bp%7D%3D%20%5Cfrac%7B1%7D%7Bp%7D%5B%20%5Cfrac%7BR%7D%7Bm%7D%2B%20%5Cfrac%7Bb%7D%7Bm%5E%7B2%7D%7D%5D%20%20%20)