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muminat
3 years ago
10

Simplify 25/26•13/15 HELPME

Mathematics
2 answers:
dezoksy [38]3 years ago
7 0
\frac{25}{26} * \frac{13}{15} = \frac{5*5}{13*2} * \frac{13}{3*5} = \frac{(13*5)*5}{(13*5)*3*2} = \frac{5}{6}
leva [86]3 years ago
6 0
\frac{25}{26} * \frac{13}{15} =  \frac{25~*~13}{26~*~15} =  \frac{325}{390} = \frac{5}{6}
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nikitadnepr [17]
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nordsb [41]

Answer:

y= 9

Step-by-step explanation:

4y-y+11=38

3y+11=38

3y= 38-11

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attashe74 [19]
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7 0
3 years ago
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Elena-2011 [213]

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8 0
3 years ago
Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations. Hint: Use
QveST [7]
\bf 27\implies 3^3\\\\
i^3\implies i\cdot i\cdot i\implies i^2\cdot i\implies -1\cdot i\implies -i\\\\
-----------------------------\\\\
x^3-27i=0\implies x^3+3^3(-i)=0\implies x^3+(3^3i^3)=0
\\\\\\
x^3+(3i)^3=0\\\\
-----------------------------\\\\
\textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3\\\\
-----------------------------\\\\

\bf (x+3i)[x^2-3ix+(3i)^2]=0\implies 
\begin{cases}
x+3i=0\implies \boxed{x=-3i}\\\\
x^2-3ix+(3i)^2=0
\end{cases}
\\\\\\
\textit{now, for the second one, we'd need the quadratic formula}
\\\\\\
x^2-3ix+(3i)^2=0\implies x^2-3ix+(3^2i^2)=0


\bf x^2-3ix+(-9)=0\implies x^2-3ix-9=0
\\\\\\
\textit{quadratic formula}\\\\
x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}\implies x=\cfrac{3i\pm\sqrt{(-3i)^2-4(1)(-9)}}{2(1)}
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x=\cfrac{3i\pm\sqrt{27}}{2}\implies \boxed{x=\cfrac{3i\pm 3\sqrt{3}}{2}}
7 0
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