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Bad White [126]
3 years ago
11

Please please help meeee

Mathematics
1 answer:
mote1985 [20]3 years ago
8 0
With what?
You gotta post something
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Determine the number of solutions to the following system of equations.
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Evaluate f (x) = 3 (2)x for x=-1
kap26 [50]

The value of the function at x = -1 is 5

<h3>What is  function?</h3>

Function can be defined as an expression, rule, or law that defines a relationship between one variable known as the independent variable and another called the dependent variable.

From the information given, we have that;

f (x) = 3 (2)x  with  x=-1

Now, let's substitute the value of x as - 1 in the function given;

f (x) = 3 (2)x

f (x) = 3 (2) -1

Expand the bracket;

f (x) = 6 - 1

f(x) = 5

Thus, the value of the function at x = -1 is 5

Learn more about functions here:

brainly.com/question/6561461

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8 0
1 year ago
Explain why sin^-1(sin(3pi/4))does not equal 3pi/4 when y=sin x and y=sin^-1 x are inverses.
Fofino [41]

Because they're not inverses, not exactly. \sin x is not invertible on its entire domain because it's not one-to-one. There are infinitely many values of x such that \sin x=0, for example.

The standard function \sin^{-1}x has a domain of -1\le x\le1 and outputs values between -\dfrac\pi2 and \dfrac\pi2. This means that its inverse, \sin x, is indeed its inverse as long as -\dfrac\pi2\le x\le\dfrac\pi2.

\dfrac{3\pi}4 is larger than \dfrac\pi2 and thus does not fall in the "invertible part" of the domain of \sin x. We have

\sin\dfrac{3\pi}4=\dfrac1{\sqrt2}

which is a value between -1 and 1, so that

\sin^{-1}\left(\sin\dfrac{3\pi}4\right)=\sin^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4

If we wanted to recover \dfrac{3\pi}4, we'd have to redefine \sin^{-1}x or define a new inverse function that works on a different branch of the domain of \sin x.

7 0
3 years ago
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