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25 Jan 2021, 07:45
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E
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Difficulty:
35%
(medium)
Question Stats:
70% (01:39) correct
30%
(01:37)
wrong
based on 192
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If \((x + 3)^2 = 16\) and \((y + 4)^2 = 36\), then what is the maximum value of \(\frac{x}{y} =\) ?
A. –7/10
B. –1/2
C. 0
D. 1/2
E. 7/10
A. –7/10
B. –1/2
C. 0
D. 1/2
E. 7/10
25 Jan 2021, 08:28
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(x + 3)^2 = 16
x+3 = +4 or -4
x = 1 or -7
(y + 4)^2 = 36
y+4 = +6 or -6
y = 2 or -10
Max of x/y is when x = -7 and y = -10
x/y = 7/10
E is the answer.
x+3 = +4 or -4
x = 1 or -7
(y + 4)^2 = 36
y+4 = +6 or -6
y = 2 or -10
Max of x/y is when x = -7 and y = -10
x/y = 7/10
E is the answer.
25 Jan 2021, 09:17
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Asked: If \((x + 3)^2 = 16\) and \((y + 4)^2 = 36\), then what is the maximum value of \(\frac{x}{y} =\) ?
x=-7 or 1
y=-10 or 2
x/y = 7/10 or 1/2 or -7/2 or -1/10
IMO E
Posted from my mobile device
x=-7 or 1
y=-10 or 2
x/y = 7/10 or 1/2 or -7/2 or -1/10
IMO E
Posted from my mobile device
