anonymous
  • anonymous
simplify each: please explain how to do it rather than just giving me the answer
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
each what?
anonymous
  • anonymous
a. |dw:1371246974503:dw| b. |dw:1371247043561:dw| c. |dw:1371247114328:dw| d. |dw:1371247190593:dw| e. |dw:1371247275341:dw|
anonymous
  • anonymous
for the first one, cross multiply first

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anonymous
  • anonymous
that also means changing the second fraction into two different ones. damn. this is kinda complicated to explain. :/
anonymous
  • anonymous
ok. first cross multiply (2u^3/b^2) with (2u^3/b^2). then with the result, cross multiply it with the first fraction
anonymous
  • anonymous
you could draw it out
anonymous
  • anonymous
|dw:1371248249989:dw|
anonymous
  • anonymous
i cant draw numbers!!!!!
anonymous
  • anonymous
if you click the A on the draw thing you can type numbers
anonymous
  • anonymous
wouldnt that just be 1?
anonymous
  • anonymous
should she do the exponents first?
anonymous
  • anonymous
can you help me @melody16
Mertsj
  • Mertsj
|dw:1371258721049:dw|
anonymous
  • anonymous
for the first one you should do the exponent outside of the parenthesis. when the exponent is outside a parenthesis you need to multiply the other exponents instead of adding it.
Mertsj
  • Mertsj
|dw:1371258811609:dw|
Mertsj
  • Mertsj
Can you get it from there?
anonymous
  • anonymous
merts got you covered.
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
do you cross multiply them ?
phi
  • phi
For your first problem you should learn these rules: when multiplying fractions you multiply top times top and bottom times bottom: example with numbers: \[ \frac{2}{3} \cdot \frac{1}{2} = \frac{2 \cdot 1 }{3 \cdot 2} \] 2nd rule: you can switch the order of multiplies, example: \[ \frac{2 \cdot 1 }{3 \cdot 2} = \frac{2 \cdot 1 }{2 \cdot 3} \] of course, you can "undo" the multiply top and bottom, so you can say \[ \frac{2 \cdot 1 }{2 \cdot 3} = \frac{2}{2} \cdot \frac{1}{3} \] next rule: anything divided by itself is 1, so \[ \frac{2}{2} \cdot \frac{1}{3} = 1 \cdot \frac{1}{3}= \frac{1}{3}\]
phi
  • phi
that is a long way to get to this idea: \[ \frac{\cancel{2}}{3} \cdot \frac{1} {\cancel{2}} = \frac{1}{3}\]
phi
  • phi
next rule: a little number in the upper right means multiply by itself that many times \[ 2^3 = 2 \cdot 2 \cdot 2 \\ a^4= a \cdot a \cdot a \cdot a \]
phi
  • phi
we can use that rule plus the one up above to simplify things: \[ \frac{a^2}{a^3} \] is short for \[ \frac{a \cdot a}{a \cdot a \cdot a} \] use the cancel idea: \[ \frac{a \cdot a}{a \cdot a \cdot a} = \frac{\cancel{a} \cdot \cancel{a}}{\cancel{a} \cdot \cancel{a} \cdot a} = \frac{1}{a}\]
phi
  • phi
for a more complicated thing: \[ \left( \frac{2u^3}{b^2}\right)^2 \] it is the same rule: multiply by itself 2 times: \[ \left( \frac{2u^3}{b^2}\right)^2 = \frac{2u^3}{b^2} \cdot \frac{2u^3}{b^2} \]
anonymous
  • anonymous
|dw:1371308977582:dw|
phi
  • phi
of course, if you do these problems enough times, you notice some short cuts: \[ u^3 \cdot u^3 = u \cdot u \cdot u \cdot u \cdot u \cdot u= u^6 \] so just add the exponents is the short way
anonymous
  • anonymous
|dw:1371309186045:dw|
phi
  • phi
yes, but you can simplify \[ \frac{b^2}{b^4} \] and the u's also
anonymous
  • anonymous
ok
phi
  • phi
use the "cancel idea" \[ \frac{a \cdot a}{a \cdot a \cdot a} = \frac{\cancel{a} \cdot \cancel{a}}{\cancel{a} \cdot \cancel{a} \cdot a} = \frac{1}{a} \]
anonymous
  • anonymous
|dw:1371309527584:dw|
phi
  • phi
yes, that looks good
phi
  • phi
for problem (b), can you re-post it. this one is getting too long
anonymous
  • anonymous
I figured out b but I'll repost c

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