Let L be the low copper alloy and H be the high copper alloy.
We need 0.15L + 0.60H = 0.42 and L + H = 100
L=100 - H
Substituting this into the second equation for L, we get:
0.15(100-H) + 0.60H = 0.42(H+L)
15 - 0.15H + 0.60H = 0.42(H+100-H)
15 + 0.45H = 0.42(100)
0.45H = 42 - 15 = 27
H = 27/.45 = 60 lbs
Back substitution, L = 40 lbs.
I hope this helps
a=-8 fraction of 7
Do you need that rewritten ?
Answer:
Step-by-step explanation:
We have been given that the center of an ellipse is (1,6). One focus of the ellipse is (-2,6). One vertex of the ellipse is (10,6).
We know that standard equation of an ellipse centered at (h,k) is in form:
, where,
a = Horizontal radius
b = Vertical radius.
Since center of the ellipse is at point (1,6) and one vertex is (10,6), so the horizontal radius would be 9 (10-1) units.
Now, we will find vertical radius using formula , where c represents focal length.
Substituting given values:
Upon substituting , and in standard form of ellipse, we will get:
Therefore, the required equation of the ellipse in standard form would be .