The value range of the independent variable in the function y = radical (x + 2) / X clearly indicates that the coefficient K of the inverse proportional function is not equal to 0. Why is the answer x > = - 2 and X ≠ 0 Value range of independent variable in function y = radical (x + 2) / X Why is the answer x > = - 2 and X ≠ 0 not x > - 2 and X ≠ 0

The value range of the independent variable in the function y = radical (x + 2) / X clearly indicates that the coefficient K of the inverse proportional function is not equal to 0. Why is the answer x > = - 2 and X ≠ 0 Value range of independent variable in function y = radical (x + 2) / X Why is the answer x > = - 2 and X ≠ 0 not x > - 2 and X ≠ 0

I don't understand
This is not an inverse function, as long as the denominator is not zero, x + 2 ≥ 0 in the root sign is OK
Does the title say that it is an inverse proportion function? It's pulled by the owner himself. The inverse proportion function y = K / X
Now this function is {(x + 2) / X} under y = root
If the part under the root sign is greater than or equal to zero and the denominator cannot be zero, find the intersection
Eccentricity formula of ellipse
The book says e = C / a = root [1 + (B / a) & sup2;]
How did you get that back there?
I thought it had something to do with trigonometric function, but I didn't deduce it
When y = K / X is an inverse proportional function, where x = 1, the independent variable increases by 2 and the function value decreases by 2 / 3, then the value of K is?
It can be seen from the meaning of the title that K / (x + 2) = Y-2 / 3
Because when x = 1, y = k, x + 2 = 1 + 2 = 3
K / 3 = K-2 / 3
The solution is: k = 1
It is known that K / (x + 2) = Y-2 / 3
Because when x equals 1, y = K
So y of the above formula is replaced by K, x + 2 = 1 + 2 = 3
K / 3 = K-2 / 3
The solution is: k = 1
If the ellipse x ^ 2 / A + y ^ 2 = 1 (a > b > 0) and the hyperbola x ^ 2 / M-Y ^ 2 / N = 1 (M > 0, n > 0) have the same focus F1 and F2,
Then: P is the intersection of hyperbola, then lpf1l * lpf2l =?
Because the ellipse and hyperbola have a common focus, and the focus of hyperbola is on the X axis, so a > 1
And A-1 = m + n
Because point P is the intersection of ellipse and hyperbola, suppose Pf1 > PF2
So Pf1 + PF2 = 2 √ a, pf1-pf2 = 2 √ M
So Pf1 = √ a + √ M. PF2 = √ a - √ M
So Pf1 * PF2 = A-M = n + 1
The inverse proportion function may be "when the independent variable is a positive number, the function value decreases with the increase of the independent variable". All the inverse proportion functions are: when the independent variable is a positive number, the function value decreases with the increase of the independent variable? Let's give an example
No
I'm asking the same question
Ellipse x ^ 2 / m ^ 2 + y ^ 2 = 1 (M > 1) and hyperbola x ^ 2 / N ^ 2-y ^ 2 = 1 (n > 0) have common focus. F1, F2 and P are their intersection points. Find s of triangle f1pf2
Find the area of triangle F1 P F2
There is a common focus C ^ 2 = m ^ 2-1 = n ^ 2 + 1m ^ 2-N ^ 2 = 2, two sides plus 2n ^ 2m ^ 2 + n ^ 2 = 2n ^ 2 + 2 to find the intersection x ^ 2 = m ^ 2 (1-y ^ 2) = n ^ 2 (y ^ 2 + 1) m ^ 2-m ^ 2Y ^ 2 = n ^ 2Y ^ 2 + n ^ 2Y ^ 2 = (m ^ 2-N ^ 2) / (m ^ 2 + n ^ 2) = 2 / (2n ^ 2 + 2) = 1 / (n ^ 2 + 1), so the height of triangle = | y | = 1 / √ (n ^ 2 + 1) bottom edge F1F2 = 2C = 2
It is known that in each quadrant of the image of the inverse scale function, the value of the function y decreases with the increase of the independent variable x, and the value of K also satisfies ≥ 2k-1. If K is an integer, the analytic expression of the inverse scale function is obtained
In each quadrant of the image of the inverse scale function, the function value y decreases with the increase of the independent variable x
k＞0
The value of K also satisfies ≥ 2k-1
k≤1
K=1
y=1/x
Let F1 and F2 be the two focal points of hyperbola X29 − y216 = 1, point p be on the hyperbola, and ∠ f1pf2 = 60 °, the area of △ f1pf2______ .
From the meaning of X29 − y216 = 1, we can get F2 (5,0), F1 (- 5,0). From the cosine theorem, we can get & nbsp; 100 = Pf12 + pf22-2pf1 ·pf2cos60 ° = (pf1-pf2) 2 + Pf1 ·pf2 = 36 + Pf1 ·pf2, ｜ Pf1 ·pf2 = 64. S △ f1pf2 = 12pf1 ·pf2sin60 ° = 12 × 64 × 32 = 163
The image of a certain function passes through the point (- 1,2), and the value of function y decreases with the increase of independent variable x. please write a function relation that meets the above conditions___ .
∵ y decreases with the increase of X, ∵ K ＜ 0. And ∵ a straight line crosses a point (- 1,2), ∵ the analytic formula is y = - 2x or y = - x + 1, etc. so the answer is: y = - 2x (the answer is not unique)
If x ^ 2 / M + y ^ 2 = 1 and hyperbola x ^ 2 / n-y ^ 2 = 1 have the same focus F1F2, and P is an intersection of two hyperbolas, then the area of triangle f1pf2 is
What's the size of "△"