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madreJ [45]
3 years ago
9

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

ther step to isolate it. Let's say we now have
3C+4D=5
2C+5D=2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?
Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers.
Physics
1 answer:
Margaret [11]3 years ago
5 0

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)

First, we isolate C from equation (2):

2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)

using this value of C from equation (3) in equation (1):

3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}

<u>D = - 0.57</u>

Put this value in equation (3), we get:

C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C =  \frac{17}{7}\\

<u>C = 2.43</u>

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At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is

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A box slides down a 31° ramp with an acceleration of 0.99 m/s2. Determine the coefficient of kinetic friction between the box an
tino4ka555 [31]

Answer:\mu=0.48

Explanation:

Given

inclination \theta =31^{\circ}

Acceleration of object=0.99 m/s^2

Now using FBD

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