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3 years ago

-5y+8x=-18 5y+2x=58 Solve for x and y

Mathematics

2 answers:

xxMikexx [17]3 years ago

7 0

Answer:

in -5y+8x=-18 and 5y+2x=58

x = 4

in -5y+8x=-18 and 5y+2x=58

y = 10

Mrac [35]3 years ago

3 0

Answer:

if this is a simultaneous equation;

Step-by-step explanation:

-5y+8x=-18 [1]

5y+2x=58 [2]

solve for5y

-5y+8x=-18

5y=58-2x [4]

-[58-2x]+8x=-18

x=4

substitute x=4 into equation 4

5y=58-2*4

x=4and y=10

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3 years ago

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murzikaleks [220]

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a) $t = \sqrt{\frac{2 *10m}{9.8 m/s^2}}=1.43s$

b) $4.9t^2 -2t -10 =0$

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Step-by-step explanation:

For this case we can use the following kinematics formulas:

$y_f = y_o + V_o t + \frac{1}{2} a t^2$

For this case we assume that the only acceleration is the gravity a = g =9.8 m/s^2. And for this case we can assume that the reference point is $y_o =0$ and the final height would be $y_f = -10 m$ since is below the initial point.

Teh acceleration would be a=-g, since the gravity is acting dowward, we assume that the initial velocity is 0, so then we have everything to replace and we got:

$-10 m = 0m + (0m/s)t -\frac{1}{2} (9.8 m/s^2) t^2$

And solving for t we got:

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$-10 m = 0m + (2m/s)t -\frac{1}{2} (9.8 m/s^2) t^2$

And we have this quadratic equation:

$-10 = 2t -4.9t^2$

$4.9t^2 -2t -10 =0$

And we can use the quadratic formula to solve it:

$t = \frac{2 \pm \sqrt{(-2)^2 -4(4.9)(-10)}}{2*4.9}$

$t =\frac{2 \pm \sqrt{200}}{9.8}$

$t_1 =1.65 s, t_2 =-1.24 s$

And since the time can't be negative the correct option would be $t=1.65 s$

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Answer:

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