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tino4ka555 [31]
3 years ago
10

2/3x=10 help please and explain

Mathematics
2 answers:
mixer [17]3 years ago
5 0
<span>\frac{2}{3x}=10
</span><span>Multiply both sides by 3x:
2=30x
</span>Divide both sides by 30:
\frac{2}{30}=x
Simplify:
x=\frac{1}{15}
Check:
<span>\frac{2}{3\times\frac{1}{15}}=10
</span><span>\frac{2}{\frac{3}{15}}=10
</span>\frac{15\times2}{3}=10
\frac{30}{3}=10
10=10
stira [4]3 years ago
3 0
 Solutions 

Simplify <span><span>2/3</span>x</span> to <span><span>2x/</span>3
</span>
<span><span><span>2x/</span>3</span>=10

</span>2) Multiply both sides by 3

<span>2x=10×3

3) </span>Simplify 10×3 to 30

<span>2x=30

4) </span>Divide both sides by <span>2

</span><span>x=30 </span>÷ <span>2

5) </span>Simplify <span>30 </span><span>÷ 2</span><span> </span><span>to</span><span> </span><span>15

</span><span><span>x=15

Answer = 15  

Also can be written as 1/15</span></span>
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Step-by-step explanation:

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8 0
4 years ago
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Natasha2012 [34]

As we know that the standard equation of circle is {\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}} , where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>

Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}

{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}

{:\implies \quad \sf 9+16=r^{2}}

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>

Now , putting values in our standard equation ;

{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}

{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}

<em>This is the required equation of </em><em>Circle</em>

Refer to the attachment as well !

8 0
2 years ago
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