Please divide and simplify

Mathematics

1 answer:

Damm [24]3 years ago

3 0

Answer:

-2 $y^{3$

Step-by-step explanation:

<h2> ${\frac{18x^{3} y^{5} }{9x^{3} y^{2} } \\$ </h2><h2>Divide the numbers=</h2><h2> $\frac{2x^{3} y^{5} }{x^{3} y^{2} }$ </h2><h2>Apply exponent rule=</h2><h2> $2y^{5-2}$ </h2><h2>Subtract numbers=</h2><h2> $-2y^{3$ </h2><h2 />

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What is the y-intercept of the equation of the line that is perpendicular to the line y =

igor_vitrenko [27]

Answer:

y=-5/3x+20

Step-by-step explanation:

Let the equation of the required line be represented as $y=mx+c$

This line is perpendicular to the line $y=\frac{3}{5}x+10$

$=>m*\frac{3}{5}=-1$

$=>m=\frac{-5}{3}$

So the equation of the required line becomes $y=\frac{-5}{3}x+c$

This line passes through the point (15.-5)

$-5=\frac{-5}{3}*15+c$

$=>c=20$

So the equation of the required line is $y=\frac{-5}{3}x+20$

Among the given options, option 4 is the correct one.

3 0

3 years ago

Kindly solve this question 6 too !!

Shkiper50 [21]

Answer:

$x=1$

Step-by-step explanation:

$~~~~~x = \dfrac{1}{2 - \dfrac 1{2-\dfrac{1}{2-x}}}\\\\\\\implies x = \dfrac{1}{2 - \dfrac 1{\dfrac{4-2x-1}{2-x}}}\\\\\\\implies x= \dfrac{1}{2 - \dfrac{2-x}{3-2x}}\\\\\\\implies x = \dfrac{1}{\dfrac{6-4x-2+x}{3-2x}}\\\\\\\implies x = \dfrac{3-2x}{4-3x}\\\\\\\implies 4x -3x^2 = 3-2x\\\\\\\implies 3x^2 -4x +3-2x = 0\\\\\\\implies 3x^2 -6x +3 = 0\\\\\\\implies 3(x^2 -2x +1) =0\\\\\\\implies x^2 -2x +1 = 0\\\\\\\implies (x-1)^2 = 0\\\\\\\implies x -1 = 0\\\\\\\implies x = 1$