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yan [13]
3 years ago
12

The equation of line GH is y = 4x − 3. Write an equation of a line perpendicular to line GH in slope-intercept form that contain

s point (2, 3).
y = 4x − 3

y = −4x − 5

11 over 4x − 3

y = negative 1 over 4x + 7 over 2
Mathematics
2 answers:
Dovator [93]3 years ago
6 0
Hello there.

<span>The equation of line GH is y = 4x − 3. Write an equation of a line perpendicular to line GH in slope-intercept form that contains point (2, 3).

</span><span>y = negative 1 over 4x + 7 over 2</span>
Novosadov [1.4K]3 years ago
3 0
Thank you for posting your question here at brainly. Feel free to ask more questions.  

The best and most correct answer among the choices provided by the question is <span>y = negative 1 over 4x + 7 over 2</span>.<span>       
    </span><span>
Hope my answer would be a great help for you.</span>
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What is the solution of -8/2y-8=5/y+4 - 7y+8/y^2-16
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Answer:

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Step-by-step explanation:

Domain:\\\\2y-8\neq0\ \wedge\ y+4\neq0\ \wedge\ y^2-16\neq0\\\\2y\neq8\ \wedge\ y\neq-4\ \wedge\ y^2\neq16\\\\y\neq4\ \wedge\ y\neq-4\ \wedge\ y\neq\pm\sqrt{16}\\\\y\neq4\ \wedge\ y\neq-4\ \wedge\ y\neq-4\ \wedge\ y\neq4\\\\\boxed{y\neq-4\ \wedge\ y\neq4}\\\\===========================

\dfrac{-8}{2y-8}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-16}\\\\\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{y^2-4^2}\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\\dfrac{-8}{2(y-4)}=\dfrac{5}{y+4}-\dfrac{7y+8}{(y-4)(y+4)}\qquad\text{multiply both sides by (-2)}\\\\\dfrac{8}{y-4}=-\dfrac{10}{y+4}+\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{add}\ \dfrac{10}{y+4}\ \text{to both sides}\\\\\dfrac{8}{y-4}+\dfrac{10}{y+4}=\dfrac{14y+16}{(y-4)(y+4)}

\dfrac{8(y+4)}{(y-4)(y+4)}+\dfrac{10(y-4)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\\\\\dfrac{8(y+4)+10(y-4)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{use the distributive property}\\\\\dfrac{8y+32+10y-40}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\qquad\text{combine like terms}\\\\\dfrac{(8y+10y)+(32-40)}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\\\\\dfrac{18y-8}{(y-4)(y+4)}=\dfrac{14y+16}{(y-4)(y+4)}\iff18y-8=14y+16\\\\18y-8=14y+16\qquad\text{subtract 14y from both sides}

4y-8=16\qquad\text{add 8 to both sides}\\\\4y=24\qquad\text{divide both sides by 4}\\\\y=8\in D

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