First find the the value of t where the curve intersects the Y-axis. This is when x = 0.
x = t^2 - 2t = 0 = t(t - 2)
So t= 0 and t = 2
dA = (0 - x)*dy .... Since the curve has negative x in this region
y = SQRT(t) and dy = [(1/2)/SQRT(t)]dt
dA = [2t - t^2][(1/2)/SQRT(t)]dt
dA = [t^(1/2) - (1/2)t^(3/2)]dt
Integrate to get: A = (2/3)t^(3/2) - (1/5)t^(5/2)
Now evaluate from t= 0 to t = 2.
Area = [(2/3)2^(3/2) - (1/5)2^(5/2)] - [0]
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Area = SQRT(2)[4/3 - 4/5]
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Area = SQRT(2)[8/15) = 0.754
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Answer:
B) -x+8y=56
Step-by-step explanation:
y=1/8x+7
Multiply by 8 to clear the fraction
8y = 8(1/8x+7)
Distribute
8y = 8*1/8x +8*7
8y = x+56
Subtract x from each side
-x+8y = x-x+56
-x+8y = 56
Answer: 5.541
Step-by-step explanation: hope it helps
Answer:
Step 1, and step 5
Step-by-step explanation:
The end of the equation has to equal 0, so c can not be -6 but instead -12.
They forgot about the 4 +/- the square root of 40 in the last step.