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3 years ago

Which points are 19 units apart?

Mathematics

2 answers:

Harrizon [31]3 years ago

4 0

Answer:

(-14, 12) and (5, 12)

Step-by-step explanation:

From inspection of the points given, we can see that all pairs of points share the same y-values.

Therefore, each pair of points are on the same horizontal line.

To find the distance between the points, simply find the difference between the x-values.

Given points: (-14, 12) and (5, 12)

⇒ Difference between x-values = 5 - (-14) = 19 units

Given points: (-5, 19) and (12, 19)

⇒ Difference between x-values = 12 - (-5) = 17 units

Given points: (-20, 5) and (1, 5)

⇒ Difference between x-values = -20 - 1 = -21 = 21 units

Given points: (14, 12) and (5, 12)

⇒ Difference between x-values = 14 - 5 = 9 units

vfiekz [6]3 years ago

3 0

√(-14-5)²+(12-12)²
√19²
√361
19

Yes

√(-5+12)²
√7²
7

√(-20-1)²
√21²
21

√(14-5)²
√9²
9

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Given f'(x) = (x-4)(4-2x), find the x-coordinate for relative maximum on the graph of f(x)

antiseptic1488 [7]

Answer:

The coordinate of the relative maximum is x=4.

Step-by-step explanation:

Given that the derivative of the function $f(x)$ is $f'(x) = (x-4)(4-2x)$ , the maxima and minima or the critical points can be found where $f'(x)=0,$ that is:

$f'(x) = (x-4)(4-2x)=0.$

The solutions to this equation are $x=4$ and $x=2.$

Now, if the second derivative $f''(x)$ for a function $f(x)$ is negative at a critical point, then the critical point is the relative maximum.

Therefore we want to see at which of the two critical points is $f''(x)$ negative. The second derivative is:

$\frac{\mathrm{d f'(x)}}{\mathrm{d}x}=f''(x)=-4(x-3).$

Now $f''(4)$ is $-4$ , and $f''(2)$ is $+4$ , therefore we deduce that the relative maxium is located at $x=4$ , because there the second derivative $f''(x)$ is negative.