Answer:
x = -2
Step-by-step explanation:
For this problem, we must simply solve for x. To do this, we will need equation operations, and the use of the distributive property.
Let's work this line by line until we have the value for x:
3{-x + (2x + 1)} = x - 1
3*-x + 3*(2x + 1) = x - 1
-3x + 6x + 3 = x - 1
3x + 3 = x - 1
2x = -4
x = -2
Now we can check our answer for x by plugging back into the original equation and see if the left hand side is equal to the right hand side:
3{-x + (2x + 1)} = x - 1
3{-(-2) + (2(-2) + 1)} ?= (-2) - 1
3{2 + (-4 + 1)} ?= -3
3{2 + (-3)} ?= -3
3{-1} ?= -3
-3 == -3
Thus, we have found the solution for x to be equivalent to negative 2.
Cheers.
Answer:
17.690 miles
Step-by-step explanation:
i took the test
Answer:
Where is the picture? I can help
Step-by-step explanation:
Answer:
Below.
Step-by-step explanation:
So calculate.
-1/6x+3+1/3x+1=x+4.
Cancel equal terms.
-1/6x+4+1/3x=x+4
Multiply both sides -1/6x+1/3x=x
collect like terms.
-x+2x=6x
move the variable from right to left.
x=6x
collect like terms
x-6x=0
Divide both sides
-5x=0
x=0
Answer:
![L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20%5Cdfrac%7B6%7D%7BS%5E2%2B1%7D%20%5B%5Csqrt%7B3%7D%20%5C%20S%20%2B1%20%5D)
Step-by-step explanation:
Given that:
recall that:
cos (A-B) = cos AcosB + sin A sin B
∴
![f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}]](https://tex.z-dn.net/?f=f%28t%29%20%3D%2012%20%5Bcos%5C%20%20t%20%5C%20%20cos%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%2B%20sin%20%5C%20t%20%20%5C%20sin%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%5D)
![f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}]](https://tex.z-dn.net/?f=f%28t%29%20%3D%2012%20%5Bcos%20%5C%20%20t%20%5C%20%5Cdfrac%7B3%7D%7B2%7D%2B%20sin%20%20%5C%20t%20%20%5C%20sin%20%5Cdfrac%7B1%7D%7B2%7D%5D)
![L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20L%20%28%206%20%5Csqrt%7B3%7D%20%5C%20cos%20%5C%20%28t%29%20%2B%206%20%5C%20sin%20%5C%20%28t%29%20%5D)
![L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%206%20%5Csqrt%7B3%7D%20%5C%20L%20%5Bcos%20%5C%20%28t%29%20%5D%20%2B%206%5C%20L%20%5B%20sin%20%5C%20%28t%29%20%5D)
