Can some one help me with these questions thank you

Mathematics

1 answer:

lisov135 [29]3 years ago

3 0

Positive and a positive, will always be a positive because theyre both the same. If two things are the same, theyre equal. ( this is from what ik,I read it as a common sense question)

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SOMEONE PLEASE HELP ME ASAP PLEASE!!!!!

Liula [17]

Answer:

T(x, y) = (x - 5, y - 6)

Step-by-step explanation:

The transformation T is a translation that creates an image P' of original point P by adding 5 to the x-coordinate and 6 to the y-coordinate of the original point P.

Now you want to transform point P' into point P, so you are undoing the first transformation. You need to do the opposite of what the transformation above does. You need to subtract 5 from the x-coordinate and subtract 6 from the y-coordinate.

Answer: T(x, y) = (x - 5, y - 6)

5 0

3 years ago

This is really confusing me:<br> Simplify so that your answers contain only positive exponents.

xxTIMURxx [149]

Answer:

see below

Step-by-step explanation:

For simplifying expressions of this sort, there are four rules of exponents that come into play;

$a^ba^c=a^{b+c}\\\\(ab)^c=a^cb^c\\\\(a^b)^c=a^{bc}\\\\\dfrac{1}{a^b}=a^{-b}\ \quad\text{which means}\quad a^b=\dfrac{1}{a^{-b}}$

I find it convenient to eliminate the fractions by adding the exponents, then rewrite any negative exponents as denominator factors.

$23. \quad\dfrac{1}{x^{-2}}=x^2\\\\24. \quad\dfrac{mn}{m^2n^3}=m^{1-2}n^{1-3}=m^{-1}n^{-2}=\dfrac{1}{mn^2}\\\\25. \quad\dfrac{k^{-2}}{k^{-3}}=k^{-2-(-3)}=k\\\\26. \quad\left(\dfrac{m^2}{n^3}\right)^3=\dfrac{m^{2\cdot 3}}{n^{3\cdot 3}}=\dfrac{m^6}{n^9}\\\\27. \quad\dfrac{x^2y^3z^4}{x^{-3}y^{-4}z^{-5}}=x^{2-(-3)}y^{3-(-4)}z^{4-(-5)}=x^5y^7z^9$