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Gala2k [10]
1 year ago
12

Let g(x) = f(x+3) for some function f(x). Is g(x) a function? From the previous problem, let the domain of f(x) be [1, 4). What

is the domain of g(x)?
Mathematics
1 answer:
padilas [110]1 year ago
7 0

The domain of g(x) is [4, 7)

<h3>How to determine the domain of g(x)?</h3>

The function is given as:

g(x) = f(x + 3)

Since f(x) is a function, then g(x) is also a function

The domain of f(x) is given as:

[1, 4)

The equivalent of these values in g(x) are

x = 1 + 3 = 4

x = 4 + 3 = 7

Hence, the domain of g(x) is [4, 7)

Read more about domain at

brainly.com/question/1770447

#SPJ1

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\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=4,\:b=2\\=\sqrt{4^2+2^2}\\Refine\\=\sqrt{20}\\\sqrt{20}=2\sqrt{5}\\=2\sqrt{5}

How to get answer for number 2: | 5-i |

\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=5,\:b=-1\\=\sqrt{5^2+\left(-1\right)^2}\\=\sqrt{26}

Number 3 how to get answer: | -3i |

\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=0,\:b=-3\\=\sqrt{0^2+\left(-3\right)^2}\\Refine\\=\sqrt{9}\\\sqrt{9}\\\mathrm{Factor\:the\:number:\:}\:9=3^2\\=\sqrt{3^2}\\\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a\\\sqrt{3^2}=3\\= 3

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