12a=x
4a+6b=10
2a+3b=5
2a-4b=12
a-2b=6
2a+3b+1=a-2b
a+5b+1=0
a=-1-5b
2(-1-5b)+3b=5
-2-10b+3b=5
-2-7b=5
-7b=7
7b=-7
b=-1
-1-5*-1=a
-1+5=a
4=a
4*12=48
12a=48
Hope this helps :)
Answer:
The answer is below
Step-by-step explanation:
The next step to solve the system, would be to divide the second row by 3 and we would be left with: 3 * R2
[1 -5 3 0 -2
0 1 -7/3 0 4/3
0 0 1 2 -2]
Then what we will do is multiply row 3 by 7/3 and then subtract it from row 2, that is, R2 - 7/3 * R3, and it would look like this:
[1 -5 3 0 -2
0 1 0 14/3 -10/3
0 0 1 2 -2]
And these would be the next two steps in the process of solving the system.
Answer:
108
Step-by-step explanation:
150x0.72=108
Answer:
The total mechanical energy is 0.712 J.
Step-by-step explanation:
A blue 573 g mobile is moving on a metal rail. If on the mobile, the spring exerts a force to the right of modulus 0.85 N and in the indicated position the kinetic energy of the mobile-spring system is 0.47 J and its elastic potential energy is 0.242 J. Determine the mechanical energy of the system mobile-spring in the position shown as indicated in the figure.
The total mechanical energy is given by the sum of the kinetic energy and the potential energy.
Kinetic energy = 0.47 J
Potential energy = 0.242 J
The total mechanical energy is
T = 0.47 + 0.242 = 0.712 J
It's important that you share the complete question. What is your goal here? Double check to ensure that you have copied the entire problem correctly.
The general equation of a circle is x^2 + y^2 = r^2. Here we know that the circle passes thru two points: (-3,2) and (1,5). Given that a third point on the circle is (-7, ? ), find the y-coordinate of this third point.
Subst. the known values (of the first point) into this equation: (-3)^2 + (2)^2 = r^2. Then 9 + 4 = 13 = r^2.
Let's check this. Assuming that the equation of this specific circle is
x^2 + y^2 = r^2 = 13, the point (1,5) must satisfy it.
(1)^2 + (5)^2 = 13 is not true, unfortunately.
(1)^2 + (5)^2 = 1 + 25 = 26 (very different from 13).
Check the original problem. If it's different from that which you have shared, share the correct version and come back here for further help.
