3 years ago

Give the domain and range of the relation.

2 answers:

6 0

I’am doing this rn it’s soooo hardddd!! Let me see if I know this one

6 0

It’s the first one because remember that domain is the x - value and the range is the y- value .

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sergiy2304 [10]

Here’s the link to the answer

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shusha [124]

all the bases and sides are all each 80° then

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2 years ago

s2008m [1.1K]

Answer:

81 squared feet

Step-by-step explanation:

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3 years ago

amid [387]

This is not true.

$\tan^2x+\sec^2x=\dfrac{\sin^2x}{\cos^2x}+\dfrac1{\cos^2x}=\dfrac{1-\cos^2x+1}{\cos^2x}=2\sec^2x-1$

$2\sec^2x-1=1\implies\sec^2x=1\implies\sec x=\pm1\implies x=n\pi$

where is $n$ is any integer. So suppose we pick some value of $x$ other than these, say $x=\dfrac\pi4$ . Then

$\tan^2\dfrac\pi4+\sec^2\dfrac\pi4=1+2=3\neq1$

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3 years ago

soldier1979 [14.2K]

Answer:

7 and 5

Step-by-step explanation:

7+5=12 and 7-5=2

8 0

2 years ago