Answer:
The answer is 12
Step-by-step explanation:
Step 1: $26.95 - $8.95 (which equals $18.00)
Step 2: 18 divided by 1.50 (which equals 12)
Answer:
<em>4 miles. </em>
<em> </em><em>Step</em><em> </em><em>by</em><em> </em><em>step</em><em> </em><em>explana</em><em>tion</em>
The example for an inconsistent system of equations:
x + 2 y = 10 and x + 2 y = 20. If we try to solve it algebraically:
x = 10 - 2 y; 10 - 2 y + 2 y = 20; 10 = 20 ( not true ). We will with something that is not true. Graph of those equations shows 2 parallel lines. Therefore the system is inconsistent.
The system is dependent when we have equivalent equations. For example:
2 x + y = 5 and 4 x + 2 y = 10. If we try to solve this system, we will end up with: 0 = 0 ( true ). The graph of this system shows the same line. This system is dependent.
Finally, if we have just one solution for the system of equations then the system is independent. For example: x + y = 5 and - x + 2 y = 10. When we add those equations: 3 y = 15; y = 15 : 3; y = 5; x + 5 = 5; x = 0. So we have just one solution ( x , y ) = ( 0, 5 ). The system of equations is independent.
64 x5-26=320-26=294. So 294
Explanation:
The Law of Cosines specifies the relationship between the three sides of a triangle and any one of the angles. If the sides are designated a, b, and c, and the angle opposite side c is C, then it tells you ...
c² = a² + b² -2ab·cos(C)
This relationship can be used to find any and all angles, given the three sides of a triangle. Or, having found one angle using the Law of Cosines, the others can be found using the Law of Sines:
sin(A)/a = sin(B)/b = sin(C)/c
_____
Typically, inverse functions are required. That is, from the Law of Cosines, ...
C = arccos((a² +b² -c²)/(2ab))
And from the Law of Sines, ...
A = arcsin(a/c·sin(C))
B = arcsin(b/c·sin(C))
_____
<em>Note on solving triangles</em>
It often works best to make use of exact values where possible. It is also a good idea to start with the longest side/largest angle. Of course, once you have two angles the other can be found as the supplement of their sum.
