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Monica [59]
3 years ago
9

Which expression is equivalent to ^3√×^5y?​

Mathematics
2 answers:
Komok [63]3 years ago
4 0

Answer:

Combine any like terms on each side of the equation: x-terms with x-terms and constants with constants. Arrange the terms in the same order, usually x-term before constants. If all of the terms in the two expressions are identical, then the two expressions are equivalent.

katovenus [111]3 years ago
4 0

Answer:

Step-by-step explanation:

x=  

    3

------------

  5y+2

​  

 

 

 

 

SOLVE FOR Y

y=  

  5

--------------

 3x−2

​  

 

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Please answer question three and four if you can :)<br> Show full working out ty;)
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Step-by-step explanation:

3

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

D = ( \frac{2 + 5}{2},  \:  \:  \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \:  \:  \frac{ 1}{2} ) \\ \therefore D= (3.5, \:  \: 0.5) \\  \& \: A=(-1,\:\:4)...(given) \\\\  now \: by \: distance \: formula \\  \\ Length  \: of \:  segment  \: AD \\  =  \sqrt{( - 1 - 3.5)^{2}  +  {(4 - 0.5)}^{2} }  \\ =  \sqrt{(4.5)^{2}  +  {(3.5)}^{2} }  \\ =  \sqrt{20.25 + 12.25 }  \\  =  \sqrt{32.5}  \\    \red{ \boxed{\therefore Length  \: of \:  segment  \: AD  = 5.7 \: units}}

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

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Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

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