Answer:
x=133 y=-25
Step-by-step explanation:
I'll do both ways for you. So let's start with Substitution:
With the sub method, you have to set both equations equal to each other by setting them equal to the same variable. Since there is no coefficient in front of both x's in both equations, that variable will be easiest to solve for.
x + 2y = 83 & x + 5y = 8
Solve for x.
x = 83 - 2y & x = 8 - 5y
Once you have solved for x, set each equation equal to one another and solve for y now.
83 - 2y = 8 - 5y
Isolate all variables to one side:
83 = 8 - 3y
Now subtract the 8 to fully isolate the y variable:
75 = -3y
Divide by -3:
-25 = y Now that you have your first variable, plug it into one of the original equations and solve for x.
x + 2(-25) = 83
x - 50 = 83
x = 133
Now for the Elimination method. For this method you need to get rid of a variable by either subtracting/adding each equation together. Again, since you can subtract and x from both equations, you will be left with only the y variable to solve:
Put each equation on top of one another and subtract:
x + 2y = 83
- (x + 5y = 8)
The x's will cancel out:
(x - x) + (2y - 5y) = (83 - 8)
Simplify:
-3y = 75
Solve for y
y = -25
Then, plug y = -25 into one of the original equations:
x + 5(-25) = 8
Solve for x:
x - 125 = 8
x = 133
Hope this helps!
I think the answer would have to be complex
Answer:
x = (5 + i sqrt(15))/4 or x = (5 - i sqrt(15))/4
Step-by-step explanation:
Solve for x:
2 x^2 - 5 x + 5 = 0
Hint: | Using the quadratic formula, solve for x.
x = (5 ± sqrt((-5)^2 - 4×2×5))/(2×2) = (5 ± sqrt(25 - 40))/4 = (5 ± sqrt(-15))/4:
x = (5 + sqrt(-15))/4 or x = (5 - sqrt(-15))/4
Hint: | Express sqrt(-15) in terms of i.
sqrt(-15) = sqrt(-1) sqrt(15) = i sqrt(15):
Answer: x = (5 + i sqrt(15))/4 or x = (5 - i sqrt(15))/4
I found a missing question online. <span>What is the magnitude of the angular displacement of the ride in radians between times t=0 and t= t1?
We can imagine our ride traveling from the starting point A to some point B (at t=1s).
We can find the angle of both points, and when we subtract them we get angular displacement.
</span>
<span>Our angular displacement is:
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