SET 1
- The question mentions that for adhering to the diet the individual can allow 300 cal for the meal
We have –
1 ounce of milk – 20 cal
1 oz of cornflakes – 160 cal
Therefore, the individual can have either:
300/20 = 15 ounces of milk
Or
300/160 = 1.875 oz of cornflakes
Say one ounce of milk = M
And 1 oz of cornflakes = C
M as a function of C = M(C) = 15/1.875 = 8C
And
C as a function of M = C(M) = 1.875/15 = 0.125M
- We are given –
T = ar2+b
We know that the units on both sides of an equation should match.
T = temperature (say it is measured in Kelvins, K)
Therefore,
a)
ar2 = K
Since r is in km, a should be in:
a =K/ r2 i.e. temp/km2
And b will be measured in temp.
b)
Let us call the 2 set of values –
T1 = ar12+b
T2 = ar22+b
Now, r1 = 1260 km r2 = 6260 km
T1 = 4300 K T2 = 1150 K
4300 = a*(1260)2 + b –(1)
1150 = a*(6260)2 + b –(2)
(1) – (2):
4300-1150 = a*(12602-62602)
3150 = (1587600 – 39187600)*a
3150 = -37600000*a
I.e., a = -8.37766 x 10-5
Putting this in (1):
We get , b = 4300 – a*(1260)2
i.e. b = 4433.00000
We generally consider the parameter with most decimal places, which in this case is a.
Thus, we should have 5 significant figures after the decimal place for the answers.
= o-b
Let us call the 2 set of values –
1 = o-b
2 = o-b
We have-
[C1] = 1.1 x 10-4 mol dm-3
[C2] = 1.58 x 10-4 mol dm-3
1 =0.016 Sm-1 2 = 0.02 Sm-1
Substituting these in the two equations –
0.016 = o -b-4 – (1)
0.02 = o -b-4 – (2)
Solving these equations simultaneously-
(2)– (1):
0.02 – 0.016 = – b (-4 – -4)
i.e. b = -1.92149 Sm-1/ mol dm-3
Solving this in (1) –
o = -0.00415277 Sm-1
- We have the following information from the problem statement-
- The equation varies sinusoidally. But since the range of a sin wave is from -1 to +1, this wave must be a scaled and shifted version of a sinusoid
Y = Asin(t+)
- Time period is = 1 minute i.e. 60 seconds T = 60 sec
i.e. f = 1/60 Hz
= 2 f = 2/T =
Therefore, we have:
The axis for this curve will be = (60+160)/2
= 110
I.e. A = Amplitude = min A – max A = 160-60 = 100
Thus, the formula for the sin wave is = 100sin(+)
5.
Let the 2 heights be h1 (height at low tide) and h2 (height at high tide)
Let the 2 times be t1 (time at low tide) and t2 (time at high tide)
We are given that-
Height h1 = 7 feet
Height h2 = 11 feet
Time t1 = 4 hours
Time t2 = 10 hours (both counted in hours since midnight)
Since, the height of the tide is a function of time,
We have a general equation-
H = at+b
Now,
7 = 4a + b – (1)
11 = 10a + b -(2)
Now,
(2) – (1):
4 = 6a
i.e. a = 4/6 = 2/3
Now, substitute a =2/3 in (1)-
7 = 4*2/3 + b
i.e. b =7 – 8/3 = 13/3
Thus, we have the final equation-
H = at+b
H = 2t/3 + 13/3
Or
3H = 2t + 13
SET 2
- g(t)= t2-3t
h(z) = 2z – 1
(a) g(h(z)) = g(2z – 1)
= (2z-1)2-3(2z-1)
= 4z2-4z + 1-6z + 3
= 4z2-10z+4
(b) h(g(t)) = h(t2-3t)
= 2(t2-3t)-1
= 2t2-6t-1
(c) h(h(t)) = h(2t-1)
= 2(2t-1) -1
=4t-2-1
= 4t-3
(d) g(g(z)) = g(z2-3z)
= (z2-3z)2– 3(z2-3z)
= z4-6z3+9z2-3z2+9z
= z4-6z3+6z2+9z
- Any number divided by zero is undefined.
Thus for all those denominators where F(x) = 0, the rules will not be valid for those values of x.
(a) Denominator = x2 – 9
For x2 – 9 =0
x could be = -3 or +3
Therefore, for x= -3 and 3 , f(x) does not apply
(b) g(x) =
Now, = 0 for x=0
Thus for x = 0, g(x) is not valid
(c) h(x) = / x
Now, x = 0 will make h(x) invalid
- (a) The straight line equation (slope intercept formula)states that-
è y-y1/x-x1 = y2-y1/x2-x1
The intercepts in this case are:
(2,-3) and (4,5)
y+3/x-2 = 8/2
à (y+3)/(x-2) = 4
y+3 = 4x – 8
i.e. y = 4x -11
m = 4 and c =-11
(b) We know that equation of a line is-
Line, y = mx +c
Slope m = -2
Y intercept c = 3
Therefore,
The equation of the given line is-
y = -2x + 3
(a) cos x = -3/2
x = 5/6 + 2n radians where n
(b) tan = 1
= /4 + n where n
(c) sin x = -1/2
x = -/6 + 2n where n
y = f(x) = 1 + 3sin(/4)
The +1 indicates, graph is shifted left by 1 unit.
A = 3 i.e. sin wave scaled by 3 times
x/4 indicates times period is quarter of a sin wave. I.e. /2
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