$17.80 in 1 hr
$1 in 1/17.80 hr
$675.00 in 675.00/17.80 hr
37.9 hours
Which is 38 hours after rounding it.
16x^2 + 25y^2 + 160x - 200y + 400 = 0 Rearrange and regroup.
(16x^2 + 160x) + (25y^2 - 200y ) = 0-400. Group the xs together and the ys together.
16(X^2 + 10x) + 25(y^2-8y) = -400. Factorising.
We are going to use completing the square method.
Coefficient of x in the first expression = 10.
Half of it = 1/2 * 10 = 5. (Note this value)
Square it = 5^2 = 25. (Note this value)
Coefficient of y in the second expression = -8.
Half of it = 1/2 * -8 = -4. (Note this value)
Square it = (-4)^2 = 16. (Note this value)
We are going to carry out a manipulation of completing the square with the values
25 and 16. By adding and substracting it.
16(X^2 + 10x) + 25(y^2-8y) = -400
16(X^2 + 10x + 25 -25) + 25(y^2-8y + 16 -16) = -400
Note that +25 - 25 = 0. +16 -16 = 0. So the equation is not altered.
16(X^2 + 10x + 25) -16(25) + 25(y^2-8y + 16) -25(16) = -400
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = -400 +16(25) + 25(16) Transferring the terms -16(25) and -25(16)
to other side of equation. And 16*25 = 400
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 25(16)
16(X^2 + 10x + 25) + 25(y^2-8y + 16) = 400
We now complete the square by using the value when coefficient was halved.
16(x-5)^2 + 25(y-4)^2 = 400
Divide both sides of the equation by 400
(16(x-5)^2)/400 + (25(y-4)^2)/400 = 400/400 Note also that, 16*25 = 400.
((x-5)^2)/25 + ((y-4)^2)/16 = 1
((x-5)^2)/(5^2) + ((y-4)^2)/(4^2) = 1
Comparing to the general format of an ellipse.
((x-h)^2)/(a^2) + ((y-k)^2)/(b^2) = 1
Coordinates of the center = (h,k).
Comparing with above (x-5) = (x - h) , h = 5.
Comparing with above (y-k) = (y - k) , k = 4.
Therefore center = (h,k) = (5,4).
Sorry the answer came a little late. Cheers.
Answer:
Option (1)
Step-by-step explanation:
Angles 2 and 5 are supplementary by the same-side interior angles theorem.
Answer:
254
Step-by-step explanation:
see the picture for how I worked this out
The velocity of a train starting from rest moves with a uniform acceleration of 5 m/s2 would be 100 m/s.
Acceleration = 5 m/s2
Distance, s = 1 km = 1000m
Intital velocity, u=0 m/s
<h3>What is the first equation of motion?</h3>
We know that, Newton's equation of motion,
s = ut+ 1/2 at^2
1000 = 0 + 1/2 x 5 x t^2
1000 = 5/2 x t^2
2000/ 5 = t^2
t = 20 s
Also,a = (v-u)/t
5 = v-0/20
100 = v
v = 100 m/s
Hence, the velocity of a train starting from rest moves with a uniform acceleration of 5 m/s2 would be 100 m/s.
Learn more about velocity here;
brainly.com/question/3491339
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