Ln x-ln(x-5)=ln2 Help in this problem please having problems on my homework
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We're given the implicit equation
x² + 2x + y² - 4y = 12
where "implicit" means that one of the variables is considered a function of the other. In this case, the variable y = y(x) depends on the value of the variable x - in other words, the possible values of y are implied by the value of x. We then say that this equation defines y as an implicit function of x.
We're also given the implicit derivative,
dy/dx = (-2x - 2)/(2y - 4)
which tells us the rate of change y as a function of both x and y. The value of dy/dx corresponds to the slope of the tangent line to the curve defined by the implicit equation above at some point (x, y).
The tangent line is horizontal when its slope is zero. This happens when
(-2x - 2)/(2y - 4) = 0
If y ≠ 2, then we can eliminate the denominator and we're left with
-2x - 2 = 0
Solve for x :
-2x = 2
x = -1
This tells us that the points on the curve with x-coordinate -1 have a tangent line to the curve that is horizontal.
The tangent line is vertical when its slope is undefined/infinite, or equivalently when the denominator of dy/dx is zero:
2y - 4 = 0
Solve for y :
2y = 4
y = 2
So any point on the curve with y-coordinate 2 will have a tangent line there that is vertical.
By completing the square in the implicit equation, we can easily identify where these points are located.
x² + 2x + y² - 4y = 12
x² + 2x + 1 + y² - 4y + 4 = 17
(x + 1)² + (y - 2)² = 17
This equation defines a circle centered at (-1, 2) with radius √17. When x = -1, we have
(y - 2)² = 17 ⇒ y = 2 ± √17
and when y = 2, we have
(x + 1)² = 17 ⇒ x = -1 ± √17
See the attached plot to see the circle and the tangents at these points.
Answer:
Variance is the squared deviation and deviation is the difference of observed value from other values
Step-by-step explanation:
Variance is the squared deviation and deviation is the difference of observed value from other values
Where,
Putting this as an arithmetic sequence gives:
The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L
The sum of an arithmetic series can be written as:
![S_n=n/2 [2a+(n-1)d] = 784 000 \\n/2[2(150)+(n-1)200] = 784 000 \\n[300+200(n-1)=1 568 000 \\300n+200n^2-200n = 1 568 000 \\200n^2+100n- 1 568 000 = 0 \\2n^2 +n- 15680 = 0 \\n= 88.2...,-88.7](https://tex.z-dn.net/?f=S_n%3Dn%2F2%20%5B2a%2B%28n-1%29d%5D%20%3D%20784%20000%0A%5C%5Cn%2F2%5B2%28150%29%2B%28n-1%29200%5D%20%3D%20784%20000%0A%5C%5Cn%5B300%2B200%28n-1%29%3D1%20568%20000%0A%5C%5C300n%2B200n%5E2-200n%20%3D%201%20568%20000%0A%5C%5C200n%5E2%2B100n-%201%20568%20000%20%3D%200%0A%5C%5C2n%5E2%20%2Bn-%2015680%20%3D%200%0A%0A%5C%5Cn%3D%2088.2...%2C-88.7)
n has to be positive, so we get
n = <u>88.2 hours (3 s.f.)</u>
The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L
The sum of an arithmetic series can be written as:
n has to be positive, so we get
n = <u>88.2 hours (3 s.f.)</u>