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saul85 [17]
4 years ago
10

Please help with another math question

Mathematics
1 answer:
Vladimir79 [104]4 years ago
8 0
Given system of equations:
2y=4x+20 Eq 1
4y=6x+24 Eq 2
Divide both sides of equation 1 by 2.
y=2x+10
Substitute value of y in equation 2.
4(2x+10)=6x+24
8x+40=6x+24
2x+40=24
2x=-16
x=-8
Put value of x in y=2x+10
y=2(-8) + 10
y=-16+10
y=-6

Answer : (-8,-6)
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A ball is thrown from an initial height of 3 meters with an initial upward velocity of 30 m/s. The ball's height h (in meters) a
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~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{1}t^2\stackrel{\stackrel{b}{\downarrow }}{-6}t\stackrel{\stackrel{c}{\downarrow }}{+2}=0 ~\hspace{10em} t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t= \cfrac{ - (-6) \pm \sqrt { (-6)^2 -4(1)(2)}}{2(1)}\implies t=\cfrac{6\pm\sqrt{36-8}}{2}

t=\cfrac{6\pm\sqrt{28}}{2}\implies t=\cfrac{6\pm\sqrt{2^2\cdot 7}}{2}\implies t=\cfrac{6\pm 2\sqrt{7}}{2}\implies t=\cfrac{2(3\pm 1\sqrt{7})}{2} \\\\\\ t=3\pm \sqrt{7}\implies t= \begin{cases} 3+\sqrt{7}\\ 3-\sqrt{7} \end{cases}\implies t\approx \begin{cases} 5.65\\ 0.35 \end{cases}

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