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vovikov84 [41]
3 years ago
12

How do you simplify this?x²y+xy² / y²+2/5 × xy​​​

Mathematics
1 answer:
polet [3.4K]3 years ago
3 0

\huge \boxed{\mathbb{QUESTION} \downarrow}

  • How do you simplify this?
  • x²y+xy² / y²+2/5 × xy

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

\sf\frac{  { x  }^{ 2  }  y+x { y  }^{ 2  }    }{  { y  }^{ 2  }  + \frac{ 2  }{ 5  }   \times  xy  } \\

Factor the expressions that are not already factored.

_____

<u>How </u><u>to</u><u> factorise</u><u> </u><u>:</u><u>-</u>

<u>NUMERATOR</u> \downarrow

\sf \: x ^ { 2 } y + x y ^ { 2 }

Factor out xy.

\sf \: xy\left(x+y\right)

<u>DENOMINATOR</u> \downarrow

\sf{ y  }^{ 2  }  + \frac{ 2  }{ 5  }   \times  xy \\

Factor out 1/5.

\sf \: {\frac{1}{5}y\left(2x+5y\right)}  \\

_____

Continuing...

\sf\frac{xy\left(x+y\right)}{\frac{1}{5}y\left(2x+5y\right)}  \\

Cancel out y in both the numerator and denominator.

\sf\frac{x\left(x+y\right)}{\frac{1}{5}\left(2x+5y\right)}  \\

Expand the expression.

\sf\frac{x^{2}+xy}{\frac{2}{5}x+y}  \\

This can further simplified to as \downarrow

=    \boxed{\boxed{\bf\frac{5x\left(x  +y\right)}{2x+5y}}}

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3. What is expressed in decimal form?<br>a. 0.58<br>b. 5.8<br>c. 1.6<br>c<br>d. 625​
irga5000 [103]

Answer:

0.58

Step-by-step explanation:

because they are the Decimals

3 0
3 years ago
An electric pole is 10m high if it's shadow is 10×√3 in length find the elevation ofthe sum
lianna [129]
First, Look at the attached picture as a reference.

Now, using Pythagoras theorem, 

AC^2=AB^2+BC^2 \\\\AC=\sqrt{AB^2+BC^2 }\\\\AC=\sqrt{10^2+(10\sqrt3)^2}\\\\ AC = \sqrt{100+300}\\\\AC=\sqrt{400}\\\\AC=20~in.

Now, lets find the elevation of the sun.

<span> </span>tan\theta =  \frac{AB}{BC} \\\\ tan\theta =  \frac{10}{10\sqrt3} \\\\ \theta = tan^{-1} \left(\frac{10}{10\sqrt3}\right) \\\\ \boxed{\theta = 30  \°}

∴ The elevation of the sun is 30°

 

6 0
3 years ago
Answer asap please
Oduvanchick [21]

Answer:

(0,1) and (-2,9)

Step-by-step explanation:

A point (x,y) lies on the graph of f(x) = (\frac{1}{3} )^{x} if it satisfies the condition y = (\frac{1}{3} )^{x}

<h3>(0,1):</h3>

        x = 0 and y = 1.

This point satisfies the required condition as

        (\frac{1}{3} )^{0} = 1

Hence, this point <u>is</u> on the graph of f(x) = (\frac{1}{3} )^{x}

<h3>(3,27):</h3>

        x = 3 and y = 27.

This point doesn't satisfy the required condition as

        (\frac{1}{3} )^{3} = \frac{1}{27} ≠ 27

Hence, this point i<u>s not</u> on the graph of f(x) = (\frac{1}{3} )^{x}

<h3>(-2,9):</h3>

        x = -2 and y = 9.

This point satisfies the required condition as

        (\frac{1}{3} )^{-2} = 3^{2} = 9

Hence, this point <u>is</u> on the graph of f(x) = (\frac{1}{3} )^{x}

<h3>(-1,-\frac{1}{3}):</h3>

        x = -1 and y = -\frac{1}{3}.

This point satisfies the required condition as

        y = (\frac{1}{3} )^{-1} = 3^{1} = 3 ≠ -\frac{1}{3}

Hence, this point <u>is not</u> on the graph of f(x) = (\frac{1}{3} )^{x}

7 0
3 years ago
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