Answer:
Use continuity to evaluate the limit.
lim 16+radical x/ radical 16+x
1+9
Consider the intervals for which the numerator and the denominator are continuous.
The numerator 16+ radical x is contintuous on the interval
[0,00)
0
The denominator V16 + I is continuous and nonzero on the interval
(16,00)
X
The interval for which the quotient is continuous is the intersection of the above intervals.
16+
Therefore, the quotient
is continuous on the interval
716 +1
[-16,0]
X
Since 2
9 is in the interval (0, 0), then fis continuous at x = 9. Therefore,
19
16 + VE
lim
179 16 +2
f(9)
00
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Step-by-step explanation:
Answer:
Step-by-step explanation:
VZ = 44-27.5 = 16.5
ZY/VY = ⅝
WX = ⅝ of VX
VX = 36×8/5 = 57.6
VW = VX - WX = 21.6 units
Answer:
8x+12
Step-by-step explanation:
2x^2+7x-4=0
2x^2+8x-x-4=0
2x(x+4)-1(x+4)=0
(2x-1)(x+4)=0
So the zeros or solutions occur when x=-4 and 1/2
You could make this super easy by letting g(x) =
and f(x) =
and then the composition would be to take g(x) and put it into the x in f(x) to give you your