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balandron [24]
3 years ago
14

Line v passes through point (6, 6) and is perpendicular to the graph of y = 3/4x – 11. Line w is parallel to line v and passes t

hrough point (–6, 10). Which is the equation of line w in slope-intercept form?
a. y = 3/4x + 2
b. y = –3/4x + 2
c. y = 4/3x + 2
d. y = –4/3x + 2
Mathematics
2 answers:
Dovator [93]3 years ago
8 0
A) y= 3/4x + 2 because I just did this
Sphinxa [80]3 years ago
7 0
A y=3/4x+2
Parallel lines have the same slope
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If x+y=12 and xy=15,find the value of (x^2+y^2)
Mice21 [21]

Answer:  x^2+y^2=\dfrac{105\pm 18\sqrt6}{2}

<u>Step-by-step explanation:</u>

EQ1:  x + y = 12     --> x = 12 - y

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    ↓     ↓         ↓

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.\  y=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\.\quad =\dfrac{-(-12)\pm \sqrt{(-12)^2-4(1)(15)}}{2(1)}\\\\\\.\quad =\dfrac{12\pm \sqrt{144-120}}{2}\\\\\\.\quad =\dfrac{12\pm \sqrt{24}}{2}\\\\\\.\quad =\dfrac{12\pm 2\sqrt{6}}{2}\\\\\\.\quad =6\pm \sqrt{6}

Now, let's solve for x:

xy=15\\\\x(6\pm\sqrt6)=15\\\\x=\dfrac{15}{6\pm\sqrt6}\\\\\\x=\dfrac{15}{6\pm\sqrt6}\bigg(\dfrac{6\pm\sqrt6}{6\pm\sqrt6}\bigg)=\dfrac{6\pm \sqrt6}{2}

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y^2=(6\pm \sqrt6)^2\quad \rightarrow \quad y^2=36\pm 12\sqrt6 +6\quad \rightarrow \quad y^2=42\pm 12\sqrt6

x^2=\bigg(\dfrac{6\pm \sqrt6}{2}\bigg)^2\quad \rightarrow \quad x^2=\dfrac{42\pm 12\sqrt6}{4}\quad \rightarrow \quad x^2=\dfrac{21\pm 6\sqrt6}{2}

                                                                                 

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