NEED HELP To which family does the function y=2^x+5 belong quadratic square root exponential logarithmic

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

NEED HELP To which family does the function y=2^x+5 belong quadratic square root exponential logarithmic

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Same question for the equation y=(x+2)^(1/2)+3 The answers can be quadratic square root exponential reciprocal I know it's not exponential
\(\Large y=2^{\color{blue}{ x}}+5\impliedby hint\)
Well, I already know what the equation is... I just don't understand how to solve :/

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

we.. you asked to which it belongs to not to solve it
seems solved already anyway
I guess I don't understand completely what the differences are.
\(\Large y=(x+2)^{\frac{1}{2}+3}?\)
well.. can't be... so I assume is hmmm \(\Large y=(x+2)^{\frac{1}{2}}+3?\)
The +3 is not connected to the exponent
Yes
\(\bf y=(x+2)^{\frac{1}{2}}+3\implies y-3=(x+2)^{\frac{1}{2}} \\ \quad \\ y-3=\sqrt{x+2}\implies (y-3)^2=(\sqrt{x+2})^2\implies (y-3)^2=x+2 \\ \quad \\ (y-3)^2-2=x\implies y^2-6y+9-2=x \\ \quad \\ y^2-6y+7=x\impliedby quadratic\)
Thank you so much! I will write this down in my notes!! :) Is the first one B?
and yes, both have an exponent, and one could say they're both exponenttials BUT the difference being, one exponent is an rational or fraction the other is not so the rational can always be expressed as a root and from there you end up with some exponent on the other side
Oooooooh
\(\Large { a^{\frac{{\color{blue} n}}{{\color{red} m}}} \implies \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad \sqrt[{\color{red} m}]{a^{\color{blue} n}}\implies a^{\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ % rational negative exponent a^{-\frac{{\color{blue} n}}{{\color{red} m}}} = \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}} \implies \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}\qquad\qquad % radical denominator \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ }\) thus

Not the answer you are looking for?

Search for more explanations.

Ask your own question