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Anestetic [448]
3 years ago
11

Which piecewise relation defines a function?

Mathematics
2 answers:
RideAnS [48]3 years ago
8 0
Answer: h(x)

Explanation:

1) f(x) is not a function because there is an ambiguity for x = 4.

x = 4 belongs to both intervals - 2 ≤ x ≤ 4 and x ≥ 4

Then you find two different possible images for x: 0 and -(2)^2 = - 4.

That makes that f(x) be not a function.

2) similar thing happens with g(x)

as per the given relation the value of g(x) for x = 2 is 4 and 4+1 = 5. Which makes that g(x) be not a function.

3) j(x) is not a function because the image of x = -4 is -3(-4) = 12 and 3.

4) h(x) is a function, because there is not any ambiguity in its definition, for every x in its domain there is only one image h(x).
laila [671]3 years ago
3 0

Answer:

Choice C on edge { h(x) }

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5m - 4n, when m=5 and n = 3.
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Using algebra, find the point at which the line k(x) = 5x - 1 intersects with the line h(x) = -3x - 1 ?
vichka [17]

The point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Given: k(x) = 5x - 1, h(x) = -3x - 1

We need to find the point(if any) at which these two lines k and h meets.

To find point of intersection(if any), we need to set the functions equal as at the point of intersection the (x, y) value will be same for both of the lines.

Therefore, k(x) = h(x)

=> 5x - 1 = -3x - 1

=> 8x = 0

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k(x=0) = 5 * 0  - 1 = -1

Hence the point at which the lines k(x) = 5x - 1 and h(x) = -3x - 1 meet is (0, -1)

Know more about "point of intersection" problems here: brainly.com/question/16929168

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1 year ago
Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
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