anonymous
  • anonymous
What is the simplified form of ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@Michele_Laino
anonymous
  • anonymous
hold please<3
Michele_Laino
  • Michele_Laino
ok!

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anonymous
  • anonymous
|dw:1437504225310:dw|
Michele_Laino
  • Michele_Laino
the common denominator of these 2 fractions: 1/x and 1/y is xy
anonymous
  • anonymous
i see
Michele_Laino
  • Michele_Laino
we can rewrite your numerator as follows: \[\frac{1}{y} - \frac{1}{x} = \frac{{x - y}}{{xy}}\]
anonymous
  • anonymous
Okay:)
Michele_Laino
  • Michele_Laino
similarly, we have: \[\large \frac{1}{y} + \frac{1}{x} = \frac{{...}}{{xy}}\] please complete
anonymous
  • anonymous
2 at the top
Michele_Laino
  • Michele_Laino
what is \[\frac{{xy}}{y}\]
anonymous
  • anonymous
x
Michele_Laino
  • Michele_Laino
perfect, so we have: \[\frac{1}{y} + \frac{1}{x} = \frac{{x \cdot 1 + ...}}{{xy}}\]
anonymous
  • anonymous
2?
Michele_Laino
  • Michele_Laino
now, what is: \[\frac{{xy}}{x}\]
anonymous
  • anonymous
y
Michele_Laino
  • Michele_Laino
correct! so we have: \[\frac{1}{y} + \frac{1}{x} = \frac{{x \cdot 1 + y \cdot 1}}{{xy}} = \frac{{x + y}}{{xy}}\]
Michele_Laino
  • Michele_Laino
then your original expression can be rewritten as follows: \[\Large \frac{{\frac{1}{y} - \frac{1}{x}}}{{\frac{1}{y} + \frac{1}{x}}} = \frac{{\frac{{x - y}}{{xy}}}}{{\frac{{x + y}}{{xy}}}}\]
anonymous
  • anonymous
can't we simplify that??
Michele_Laino
  • Michele_Laino
yes! sure we can write the next step as the multiplication of the first fraction, by the inverse of the second fraction, namely: \[\Large \frac{{\frac{1}{y} - \frac{1}{x}}}{{\frac{1}{y} + \frac{1}{x}}} = \frac{{\frac{{x - y}}{{xy}}}}{{\frac{{x + y}}{{xy}}}} = \frac{{x - y}}{{xy}} \cdot \frac{{xy}}{{x + y}} = ...\]
anonymous
  • anonymous
x^2-y^2??
Michele_Laino
  • Michele_Laino
hint:|dw:1437504964076:dw|
anonymous
  • anonymous
x-y/x+y
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
i need 1 more?
Michele_Laino
  • Michele_Laino
ok! :)
anonymous
  • anonymous
Using a directrix of y = 5 and a focus of (4, 1), what quadratic function is created?
anonymous
  • anonymous
thanks!
Michele_Laino
  • Michele_Laino
that is a parabola whose equation is like this: \[y = a{x^2} + bx + c\]
Michele_Laino
  • Michele_Laino
we have to determina the corfficients a, b, and c
Michele_Laino
  • Michele_Laino
determine*
anonymous
  • anonymous
ok
anonymous
  • anonymous
so then what would I do?? Plug in?
Michele_Laino
  • Michele_Laino
the coordinates of the focus, as a function of a, b, and c, are: \[F = \left( { - \frac{b}{{2a}},\frac{{1 - {b^2} + 4ac}}{{4a}}} \right)\]
Michele_Laino
  • Michele_Laino
so we can write: \[\begin{gathered} - \frac{b}{{2a}} = 4 \hfill \\ \hfill \\ \frac{{1 - {b^2} + 4ac}}{{4a}} = 1 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
Ok :)
anonymous
  • anonymous
Where would the (4,1) go?
Michele_Laino
  • Michele_Laino
the directrix is: \[y = - \frac{{1 + {b^2} - 4ac}}{{4a}}\]
Michele_Laino
  • Michele_Laino
so we can write, by comparison: \[ - \frac{{1 + {b^2} - 4ac}}{{4a}} = 5\]
anonymous
  • anonymous
ok :)
anonymous
  • anonymous
now what would we do?
Michele_Laino
  • Michele_Laino
we have to solve this algebraic system: \[\left\{ \begin{gathered} - \frac{{1 + {b^2} - 4ac}}{{4a}} = 5 \hfill \\ - \frac{b}{{2a}} = 4 \hfill \\ \hfill \\ \frac{{1 - {b^2} + 4ac}}{{4a}} = 1 \hfill \\ \end{gathered} \right.\]
anonymous
  • anonymous
Ohhh I understand!
Michele_Laino
  • Michele_Laino
I consider the first and the third equations
anonymous
  • anonymous
would thr answer be 1/8(x+4)^2-3
Michele_Laino
  • Michele_Laino
I don't know
anonymous
  • anonymous
ahh :( It's fine
Michele_Laino
  • Michele_Laino
If I multiply both sides of the first equation by -4a, I get: \[1 + {b^2} - 4ac = - 20a\]
Michele_Laino
  • Michele_Laino
whereas if I multiply both sides of the third equation, by 4a, I get: \[1 - {b^2} + 4ac = 4a\]
Michele_Laino
  • Michele_Laino
If I sum those 2 equations together, I get: \[2 = - 16a\]
Michele_Laino
  • Michele_Laino
am I right?
anonymous
  • anonymous
yup! I believe in you :)
Michele_Laino
  • Michele_Laino
lol
Michele_Laino
  • Michele_Laino
ok! :)
Michele_Laino
  • Michele_Laino
what is a?
anonymous
  • anonymous
-8
Michele_Laino
  • Michele_Laino
hint: |dw:1437506094548:dw|
Michele_Laino
  • Michele_Laino
so, a=...?
anonymous
  • anonymous
1/-8
Michele_Laino
  • Michele_Laino
correct!
Michele_Laino
  • Michele_Laino
now, I take the second equation: \[ - \frac{b}{{2a}} = 4\]
Michele_Laino
  • Michele_Laino
Multiplying both sides by -2a, I can rewrite it as follows: \[b = - 8a\]
Michele_Laino
  • Michele_Laino
substituting a, we get: \[b = \left( { - 8} \right) \times \left( { - \frac{1}{8}} \right) = ...\]
anonymous
  • anonymous
1
Michele_Laino
  • Michele_Laino
correct! so we have: a= -1/8, b=1
Michele_Laino
  • Michele_Laino
now I take the third equation: \[\frac{{1 - {b^2} + 4ac}}{{4a}} = 1\]
anonymous
  • anonymous
|dw:1437506420592:dw|
anonymous
  • anonymous
Is that the answer? :)
Michele_Laino
  • Michele_Laino
I don't know, since I solve the question now
anonymous
  • anonymous
OK:) i'LL wait
Michele_Laino
  • Michele_Laino
I substitute into the third equation, these values: a= -1/8, and b=1, so I get: \[\frac{{1 - {1^2} + 4 \times \left( { - \frac{1}{8}} \right)c}}{{4 \times \left( { - \frac{1}{8}} \right)}} = 1\]
Michele_Laino
  • Michele_Laino
what is c?
anonymous
  • anonymous
hmmm 4?
Michele_Laino
  • Michele_Laino
I got: c=1
anonymous
  • anonymous
|dw:1437506785539:dw|
Michele_Laino
  • Michele_Laino
I don't understand the second fraction
anonymous
  • anonymous
6Y^2/ 5X^4
Michele_Laino
  • Michele_Laino
we can make these simplifications: |dw:1437506923714:dw|
anonymous
  • anonymous
so 3x^2/10y3 times 3/1
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
yay!
anonymous
  • anonymous
would the answer be 9x^2/10y^3
Michele_Laino
  • Michele_Laino
yes! correct!
anonymous
  • anonymous
last one?
Michele_Laino
  • Michele_Laino
ok! :)
anonymous
  • anonymous
i got c
Michele_Laino
  • Michele_Laino
that's right, since we can write our polynomial as below: \[f\left( x \right) = \left( {x + 7} \right)q\left( x \right)\] where q(x) is a appropriate polynomial
Michele_Laino
  • Michele_Laino
we have to solve this equation: \[{x^2} + 2x - 15 = 0\]
anonymous
  • anonymous
so then you get (x-5)(x+3)
Michele_Laino
  • Michele_Laino
the solutions are: \[\begin{gathered} x = \frac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times \left( { - 15} \right)} }}{{2 \times 1}} = \hfill \\ \hfill \\ = \frac{{ - 2 \pm \sqrt {4 + 60} }}{2} = \frac{{ - 2 \pm 8}}{2} = \begin{array}{*{20}{c}} {\frac{{ - 2 - 8}}{2} = - 5} \\ {\frac{{ - 2 + 8}}{2} = 3} \end{array} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
so would those be the discontuiny?
Michele_Laino
  • Michele_Laino
the factorization is: \[{x^2} + 2x - 15 = \left( {x + 5} \right)\left( {x - 3} \right)\]
Michele_Laino
  • Michele_Laino
yes! your discontinuities are located at x=-5 and x=3
anonymous
  • anonymous
Which polynomial identity will prove that 16 = 25 − 9?
anonymous
  • anonymous
Difference of Squares Difference of Cubes Sum of Cubes Square of a Binomial
anonymous
  • anonymous
is it A?
Michele_Laino
  • Michele_Laino
I think difference of squares, since I can write: \[\left( {x - y} \right)\left( {x + y} \right) = {x^2} - {y^2},\quad x = 5,y = 3\]
anonymous
  • anonymous
no. For the equation 16= 25-9
Michele_Laino
  • Michele_Laino
yes! please substitute x=5 and y=3, what do you get?
anonymous
  • anonymous
sub. into what?
Michele_Laino
  • Michele_Laino
hint: we have the subsequent steps: \[\begin{gathered} \left( {5 - 3} \right)\left( {5 + 3} \right) = {5^2} - {3^2} \hfill \\ 2 \times 8 = 25 - 9 \hfill \\ 16 = 25 - 9 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
Oh! So it IS A
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
yay! Solve 3x2 + 13x = 10.
anonymous
  • anonymous
i got 2/3 and -5
Michele_Laino
  • Michele_Laino
I rewrite your equation as follows: \[3{x^2} + 13x - 10 = 0\]
anonymous
  • anonymous
ok:)
anonymous
  • anonymous
use the quad. formula
Michele_Laino
  • Michele_Laino
so I get: \[\begin{gathered} x = \frac{{ - 13 \pm \sqrt {169 + 4 \times 3 \times 10} }}{{2 \times 3}} = \hfill \\ \hfill \\ = \frac{{ - 13 \pm \sqrt {289} }}{6} = \begin{array}{*{20}{c}} {\frac{{ - 13 + 17}}{2} = 2} \\ {\frac{{ - 13 - 17}}{2} = - 15} \end{array} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
Thanks!!!!
Michele_Laino
  • Michele_Laino
oops.. I have made an error
Michele_Laino
  • Michele_Laino
\[\begin{gathered} x = \frac{{ - 13 \pm \sqrt {169 + 4 \times 3 \times 10} }}{{2 \times 3}} = \hfill \\ \hfill \\ = \frac{{ - 13 \pm \sqrt {289} }}{6} = \begin{array}{*{20}{c}} {\frac{{ - 13 + 17}}{6} = \frac{2}{3}} \\ {\frac{{ - 13 - 17}}{6} = - 5} \end{array} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
just factor instead of using the formula bc that diddn't work for me
anonymous
  • anonymous
yayyy!!!!
anonymous
  • anonymous
What are the zeros of the polynomial function f(x) = x3 + x2 − 20x?
Michele_Laino
  • Michele_Laino
we have to factor out x, so we can write: \[{x^3} + {x^2} - 20x = x\left( {{x^2} + x - 20} \right)\]
Michele_Laino
  • Michele_Laino
now I solve the quadratic equation with the standard formula: \[\begin{gathered} {x^2} + x - 20 = 0 \hfill \\ x = \frac{{ - 1 \pm \sqrt {1 + 4 \times 1 \times 20} }}{{2 \times 1}} = \frac{{ - 1 \pm 9}}{2} = \hfill \\ \hfill \\ = \begin{array}{*{20}{c}} {\frac{{ - 1 - 9}}{2} = - 5} \\ {\frac{{ - 1 + 9}}{2} = 4} \end{array} \hfill \\ \end{gathered} \]
anonymous
  • anonymous
so 0,-5,4
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
|dw:1437508513154:dw|
Michele_Laino
  • Michele_Laino
the common denominator is: (x-5)(x+5)
Michele_Laino
  • Michele_Laino
so we can simplify your expression as follows: \[\frac{{\left( {4x + 2} \right)\left( {x + 5} \right) - \left( {3x - 1} \right)\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} = ...\] please complete my computation
anonymous
  • anonymous
1x+1/(x+5)(X-5)
Michele_Laino
  • Michele_Laino
I got a different result
anonymous
  • anonymous
expalin please
Michele_Laino
  • Michele_Laino
we can write this: \[\begin{gathered} \frac{{\left( {4x + 2} \right)\left( {x + 5} \right) - \left( {3x - 1} \right)\left( {x - 5} \right)}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} = \hfill \\ \hfill \\ = \frac{{4{x^2} + 22x + 10 - 3{x^2} + 16x - 5}}{{\left( {x + 5} \right)\left( {x - 5} \right)}} = ... \hfill \\ \end{gathered} \]
anonymous
  • anonymous
x^2+38x+5/(x-5)(x+5)
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
so now what do we do?
Michele_Laino
  • Michele_Laino
I think that we have completed our exercise
anonymous
  • anonymous
KK :) tHANKS SO MUCH FOR ALL YOU HELP!!!!!!! <3 <3 <3
Michele_Laino
  • Michele_Laino
:)
anonymous
  • anonymous
<3 dOZENS OF HEARTS BECAUSE YOU'RE AWESOME!!!!!!! :D
Michele_Laino
  • Michele_Laino
thanks! :)

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