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Bond [772]
3 years ago
5

Which relation is a function?

Mathematics
1 answer:
Step2247 [10]3 years ago
4 0
{(2, 3), (1, 5), (2, 7)} - NOT because 2->3 and 2->7
{(11, 9), (11, 5), (9, 3)} - NOT because 11->9 and 11->5
{(3, 8), (0, 8), (3, -2)} - NOT because 3->8 and 3->-2

{(-1, 5), (-2, 6), (-3, 7)} - YES
You might be interested in
Given the graph of a line y=−x. Write an equation of a line which is perpendicular and goes through the point (8,2).
coldgirl [10]

Answer:

y = x - 6

Step-by-step explanation:

The slope of y = -x is -1, so the slope of any line perpendicular to y = -x is +1.  Thus,

y = mx + b becomes 2 = 1(8) + b, so that b = -6.

The desired equation is y = x - 6.

Check:  Does (8,2) lie on this line?  Is 2 = 8 - 6 true?  YES.


6 0
3 years ago
Read 2 more answers
What are the answers for
Elden [556K]

Answer:

7. 589.0

8. 122.0

9. 0.6

10. 3.5

6 0
2 years ago
Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x)= 6x^1/3 + 3x^4/3. You must justify
irina [24]

Answer:

x-coordinates of relative extrema = \frac{-1}{2}

x-coordinates of the inflexion points are 0, 1

Step-by-step explanation:

f(x)=6x^{\frac{1}{3}}+3x^{\frac{4}{3}}

Differentiate with respect to x

f'(x)=6\left ( \frac{1}{3} \right )x^{\frac{-2}{3}}+3\left ( \frac{4}{3} \right )x^{\frac{1}{3}}=\frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}

f'(x)=0\Rightarrow \frac{2}{x^{\frac{2}{3}}}+4x^{\frac{1}{3}}=0\Rightarrow x=\frac{-1}{2}

Differentiate f'(x) with respect to x

f''(x)=2\left ( \frac{-2}{3} \right )x^{\frac{-5}{3}}+\frac{4}{3}x^{\frac{-2}{3}}=\frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}\\f''(x)=0\Rightarrow \frac{-4x^{\frac{2}{3}}+4x^{\frac{5}{3}}}{3x^{\frac{2}{3}}x^{\frac{5}{3}}}=0\Rightarrow x=1

At x = \frac{-1}{2},

f''\left ( \frac{-1}{2} \right )=\frac{4\left ( -1+4\left ( \frac{-1}{2} \right ) \right )}{3\left ( \frac{-1}{2} \right )^{\frac{5}{3}}}>0

We know that if f''(a)>0 then x = a is a point of minima.

So, x=\frac{-1}{2} is a point of minima.

For inflexion points:

Inflexion points are the points at which f''(x) = 0 or f''(x) is not defined.

So, x-coordinates of the inflexion points are 0, 1

7 0
3 years ago
Hexagon DEFGHI is translated on the coordinate plane below to create hexagon D′E′F′G′H′I′:
dalvyx [7]

Answer:

A

Step-by-step explanation:

it goes down by 9 and left by 3

7 0
2 years ago
ABDF is a regular hexagon. What is the measure of angle A?
dolphi86 [110]

720 \div 6 = 120
4 0
3 years ago
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