67a(t+z)=45z+67 solve for z. I need help. Please?

2 answers:

vladimir2022 [97]3 years ago

5 0

Hello there, let's get started.
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The equation is 67a(t+z)=45z+67.

Let's first distribute "67a" to the "t" and "z" in the parenthesis.
The equation simplifies to: 67at + 67az = 45z + 67.

Step one is to </span>Add -45z to both sides. 67at+67az+−45z=45z+67+−45z (this crosses 45z out), it is now simplified to 67at+67az−45z=67.

Step 2 is to now add -67at to both sides. 67at+67az−45z+−67at (this crosses 67at out)=67+−67a. The equation is now simplified to 67az−45z=−67at+67t.

Step 3: Factor out variable z. z(67a−45)=−67at+67.

Step 4: Divide both sides by 67a-45. z(67a−45)/67a−45=−67at+67/67a−45.
It's then simplified to: z=−67at+67/67a−45.

The equation: z=−67at+67/67a−45 is the final answer. Sorry if this is confusing.

xeze [42]3 years ago

4 0

67a(t+z)=45z+67
Distribute the 67a to inside the parenthesis.
67at + 67az = 45z + 67
Subtract 67 to the other side.
67at + 67az - 67 = 45z
Divide everything by 45.
1.49at + 1.49az - 1.49 = z

Hope that helped! :)

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find the magnitude of two forces f1 and f2 such that at right-angle to each other, their resultant is √13N. Also, if the two for

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This can be solved a couple of ways. One way is to use the Pythagorean theorem to write equations for the magnitude from the components of the forces. That is what was done in the graph here.

Another way is to use the Law of Cosines, which lets you make direct use of the angle between the vectors.
.. 13 = a^2 +b^2 -2ab*cos(90°)
.. 19 = a^2 +b^2 -2ab*cos(120°)
Subtracting the first equation from the second, we have
.. 6 = -2ab*cos(120°)
.. ab = 6

Substituting this into the first equation, we have
.. 13 = a^2 +(6/a)^2
.. a^4 -13a^2 +36 = 0
.. (a^2 -9)(a^2 -4) = 0
.. a = ±3 or ±2

The magnitudes of the two forces are 2N and 3N, in no particular order.