If ƒ = {(2, 5), (3, 2) (4, 6), (5, 1), (7, 2)}, then ƒ is a function. True False

Mathematics

2 answers:

Serga [27]3 years ago

7 0

True I think.......................................

Leni [432]3 years ago

6 0

Answer: True. f is a function.

Step-by-step explanation:

We know that function is a relation from a set of inputs to a set of possible outputs where each input is associated to exactly one output.

It means a relation for which each value from the set the first coordinate of the ordered pairs is related with exactly one value from the set of second coordinate of the ordered pair.

For the given set f= {(2, 5), (3, 2) (4, 6), (5, 1), (7, 2)}, here each input is associated to exactly one output.

Therefore, it is a function.

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Solve using algebraic equation: 5sin2x=3cosx (No exponents)

DaniilM [7]

$5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x$

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

$10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0$

Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$ , so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$ . More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$ , the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$ .