All categories

3 years ago

Is {(3,4),(3,5),(4,4), (4,5)} a function

Mathematics

2 answers:

Sholpan [36]3 years ago

3 0

Answer:

No, it's not a function.

Step-by-step explanation:

This is not a function because the x-inputs repeat. It would be a function if each x-input was a different number, but the inputs 3 and 4 repeat twice. Therefore, this is not a function.

Dmitry_Shevchenko [17]3 years ago

3 0

Answer:

This is NOT a function because you CANNOT have same x values twice.

Step-by-step explanation:

Hope this helps! Have a good day :)

You might be interested in

what is the median number of car show this month Frank 25, Sarah 20, randall 15, Brahim 22, Alice 20,

Tcecarenko [31]

Answer:

The median is 20.

Step-by-step explanation:

When lined from least to most (15, 20, 20, 22, 25) 20 is the middle number.

4 0

3 years ago

What is the value of x? Enter your answer in the box. units

sineoko [7]

Answer:

x = 28

Step-by-step explanation:

If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

Then x/(x +10) = 42/(42 +15)

x(57) = 42(x + 10)

57x = 42x + 420

15x = 420

x = 28

8 0

3 years ago

I’m not asking for much :(

Vinvika [58]

Answer:

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$