Well as x can never actually be -1 because I'm in the denominator -1 + 1 = 0 and we cannot divide by zero. But we can look at what number it approaches and i assume that is the relative value. sometimes functions will have asymptotes and others will have holes in the graph. this one would have an asymptote going down at a rapid rate. the asymptote would go on forever getting infinitely close to -1 but never touching. So I would say since the asymptote goes down forever that the graph approaches negative infinity
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]
</span><span>
Area = SQRT(2)[8/15) = 0.754
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Step-by-step explanation:
a×10⁰×a
<em>Since,</em><em> </em><em>m⁰</em><em>=</em><em>1</em><em> </em><em>therefore</em>
a×1×a
a²
3 x + 17 w - 18 is the answer
Solving <span>8x-5y=10 for y helps us to identify the y-intercept:
-5y = -8x + 10. Dividing both sides by -5, we get (8/5)x -2. Therefore, y = (8/5)x - 2; the y-intercept is (0,-2).
The equation </span><span>-6x-7y=-6 can be solved for its slope in the same manner.
7y = -6x + 6; then y = (-6/7)x + 6/7. Its slope is -7/6. A line perpendicular to this line has slope equal to the negative reciprocal of -7/6, which is 6/7.
So, using the slope-intercept form, y = mx + b becomes y = (6/7)x -2.</span>