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Korvikt [17]
3 years ago
12

1 1/3 Divided by 5/6

Mathematics
2 answers:
Ipatiy [6.2K]3 years ago
4 0
The answer I got is 4 2/5
UNO [17]3 years ago
3 0
1 1/3 /
5/6
=
8/5 or 1 3/5
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Simplify the expression. Rewrite using only positive exponents.
ivanzaharov [21]

The simplification form of the provided expression is x¹⁸a³b³ option

(d) x¹⁸a³b³ is correct.

<h3>What is an integer exponent?</h3>

In mathematics, integer exponents are exponents that should be integers. It may be a positive or negative number. In this situation, the positive integer exponents determine the number of times the base number should be multiplied by itself.

It is given that:

The expression:

= \rm (\dfrac{x^4a^2b^3}{x^{-2}ab^2})^3

After using the integer exponent property:

\rm =\dfrac{x^{12}a^6b^9x^6}{a^3b^6}

= a³b³x¹⁸ Or

=x¹⁸a³b³

Thus, the simplification form of the provided expression is x¹⁸a³b³ option

(d) x¹⁸a³b³ is correct.

Learn more about the integer exponent here:

brainly.com/question/4533599

#SPJ1

4 0
2 years ago
May I please have help I need it!
r-ruslan [8.4K]

Answer:

which question are you trying to answer

Step-by-step explanation:

3 0
3 years ago
Solve for the missing angle
Kobotan [32]
The missing angle is also 111 since they’re opposite interior angles (and are equal).
7 0
3 years ago
An experiment consists of drawing 1 card from a standard​ 52-card deck. Let E be the event that the card drawn is a 10. Find​ P(
viva [34]

The are 4 different possibilities to draw a ten, namely a ten of the four different suits. The probability of the event is the ratio of the 4 possibilities and the total number of possible choices, whcih is 52:

P(E) = 4/52 = 1/13 or about 7.7%

6 0
3 years ago
Find the equation of ellipse passing throgh (1,4) and (-3,2)​
irinina [24]

Answer:

\displaystyle  \frac{  {3x}^{2} }{ 35 }  +  \frac{{2y}^{2} }{  35  }   = 1

Step-by-step explanation:

we want to figure out the ellipse equation which passes through <u>(</u><u>1</u><u>,</u><u>4</u><u>)</u><u> </u>and <u>(</u><u>-</u><u>3</u><u>,</u><u>2</u><u>)</u>

the standard form of ellipse equation is given by:

\displaystyle  \frac{(x - h {)}^{2} }{ {a}^{2} }  +  \frac{(y - k {)}^{2} }{ {b}^{2} }  = 1

where:

  • (h,k) is the centre
  • a is the horizontal redius
  • b is the vertical radius

since the centre of the equation is not mentioned, we'd assume it (0,0) therefore our equation will be:

\displaystyle  \frac{  {x}^{2} }{ {a}^{2} }  +  \frac{{y}^{2} }{ {b}^{2} }  = 1

substituting the value of x and y from the point (1,4),we'd acquire:

\displaystyle  \frac{ 1}{ {a}^{2} }  +  \frac{16}{ {b}^{2} }  = 1

similarly using the point (-3,2), we'd obtain:

\displaystyle  \frac{ 9}{ {a}^{2} }  +  \frac{4 }{ {b}^{2} }  = 1

let 1/a² and 1/b² be q and p respectively and transform the equation:

\displaystyle  \begin{cases} q  +  16p  = 1  \\ 9q + 4p = 1 \end{cases}

solving the system of linear equation will yield:

\displaystyle  \begin{cases} q   =  \dfrac{3}{35} \\ \\  p =  \dfrac{2}{35}  \end{cases}

substitute back:

\displaystyle  \begin{cases}  \dfrac{1}{ {a}^{2} }   =  \dfrac{3}{35} \\ \\   \dfrac{1}{ {b}^{2} }  =  \dfrac{2}{35}  \end{cases}

divide both equation by 1 which yields:

\displaystyle  \begin{cases}  {a}^{2}   =  \dfrac{35}{ 3} \\ \\    {b}^{2}   =  \dfrac{35}{2}  \end{cases}

substitute the value of a² and b² in the ellipse equation , thus:

\displaystyle  \frac{  {x}^{2} }{  \dfrac{35}{3}  }  +  \frac{{y}^{2} }{  \dfrac{35}{2}  }   = 1

simplify complex fraction:

\displaystyle  \frac{  {3x}^{2} }{ 35 }  +  \frac{{2y}^{2} }{  35  }   = 1

and we're done!

(refer the attachment as well)

8 0
3 years ago
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