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xz_007 [3.2K]
3 years ago
15

7. Look at the picture.

Mathematics
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer:

A

Step-by-step explanation:

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Given f(x) = x2-2 and g(x) = sqrt x find f(g(2))
Ugo [173]

Answer:

\boxed{\boxed{f\bigg(g(2)\bigg)=0}}

Step-by-step explanation:

f(x)=x^2-2,\ g(x)=\sqrt{x}\\\\Domain:\\D_f:x\in\mathbb{R}\\\\D_g:x\geq0\\\\f(g(x))=\left(\sqrt{x}\right)^2-2=x-2

Used:\\\\(\sqrt{a})^2=a

f(g(2)):\\\text{Put x = 2 to the equation of the function:}\\\\f(g(2))=2-2=0

8 0
3 years ago
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Suppose that in a random sample of 240 smartphone users, 160 used an iPhone. Conduct a hypothesis test at significance level 0.0
My name is Ann [436]

a 230 metersa 230 metersa 230 metersa 230 metersa 230 metersa 230 metersa 230 metersa 230 metersa 230 meters

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2 years ago
How to Solve the inequality x+2<8
zaharov [31]

Answer:

x<6

Step-by-step explanation:

4 0
3 years ago
Simplify this expression
Pachacha [2.7K]

Answer:

\frac{1}{2} + \frac{g}{2h}

another equivalent form would be the following:

\frac{g}{2(g+h)}

Step-by-step explanation:

\frac{4g}{8(g+h)}=\frac{g}{2(g+h)} = \frac{g}{2g+2h} = \frac{1}{2} + \frac{g}{2h}


6 0
3 years ago
What best describes the asymptote of an exponential function of the form f(x)=b^x?
horrorfan [7]

It depends on the value of b


Exponential functions are defined for positive values of the base [ŧex] b [/tex]. Also, they're not defined if b=1, or at least let's say that this is a trivial case, since it is the constant function 1.


Case 0<b<1:

If the base sits between 0 and 1, the exponential function is constantly decreasing. The explanation is quite intuitive, assume for the sake of simpleness that b is rational. If it sits between 0 and 1, it means that it can be written as a fraction p/q, with p<q. So, if you give a large exponent to b, you obtain


\frac{p^x}{q^x},\quad p^x \ll q^x \text{ as } x \to \infty


On the other hand, if you consider negative exponents, you switch numerator and denominator and then raise to the same exponent the fraction q/p, which gets larger and larger.


So, if 0<b<1, we have


\lim_{x \to -\infty} b^x = \infty,\qquad \lim_{x \to \infty} b^x = 0


and thus 0 is a horizontal asymptote as x tends to (positive) infinity.


Case b>1:

This case is very similar, except all roles are inverted. Now you start with a fraction p/q where p>q. So, with positive, large exponents you get


\frac{p^x}{q^x},\quad p^x \gg q^x \text{ as } x \to \infty


And as before, negative exponents switch numerator and denominator, so the fraction becomes q/p and thus you have


\lim_{x \to -\infty} b^x = 0,\qquad \lim_{x \to \infty} b^x = \infty


So, again, 0 is a horizontal asymptote, but this time for x tending towards negative infinite.

6 0
3 years ago
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