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White raven [17]
3 years ago
13

How do you determine what could represent a function in algebra

Mathematics
2 answers:
n200080 [17]3 years ago
7 0
Hi, here is my answer:
First, if you want to determine that's relation is function, you have to realize that your value of x(Domain) is not repeated.= so, it's a function
But...If your value of x (Domain) is repeated twice or more, so I can guaranteed that is totally NOT a function

alisha [4.7K]3 years ago
4 0
Hi there! To answer your question or more questions like these you'd want to determine that relation is function, the value of x which would be your base or domain is not repeated, so by using this guideline you can determine if it's a function. On the other hand, if your value of x is repeated twice or more (+), it isn't a function. Hope this helps you and remember to mark as brainliest! BTW I like the question. It made me think a lot. :)
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<span>f(x) = -6x +6
to find inverse
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6y = 6 - x
y = 1 - x/6

Answer
</span>f^–1(x)= 1 - x/6
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Sin0=12/37 find tan0
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Answer:

\large{ \tt{❃ \: EXPLANATION}} :

  • We're provide - Sin θ = \frac{12}{37} which means 12 is the perpendicular & 37 is the hypotenuse [ Since Sin θ = \tt{  \frac{p}{h}} ] . We're asked to find out tan θ ].

\large{ \tt{❁ \: USING \: PYTHAGORAS \: THEOREM}} :

\large{ \tt{❊ \:  {h}^{2}  =  {p}^{2}  +  {b}^{2} }}

\large{ \tt{⇢ {p}^{2} +  {b}^{2}   =  {h}^{2} }}

\large{ \tt{⇢ \:  {b}^{2}  =  {h}^{2}  -  {p}^{2} }}

\large{ \tt{⇢ \:  {b}^{2} = {37}^{2} -  {12}^{2}   }}

\large{ \tt{⇢ \:  {b}^{2}  = 1369 - 144}}

\large{ \tt{ ⇢{b}^{2}  = 1225}}

\large{ \tt{⇢ \: b =  \sqrt{1225}}}

\large{ \tt{⇢ \: b = 35  \: \text {units}}}

  • Now , We know - Tan θ= \tt{ \frac{perpendicular}{base} }. Just plug the values :

\large{ \tt{➝ \: Tan  \: \theta =  \frac{p}{b}  = \boxed{ \tt{  \frac{12}{35} }}}}

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5 0
3 years ago
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