we need to take half tabulate among the 4 tablets. den it ll be like 1 tabulate is of fever and one tabulate is of cough
The first number is 10 and the second number is 5
X filling rate is 1/18 or 4/72 per hour
Y filling rate is 1/24 or 3/72 per hour
Combined as a pair it’s 7/72 in a complete 2 hour alternating sequence
72+ 7/72 =10 sequences = 70/72 on 20 hours, The outstanding balance is just 2/72
So at 20.5 hours with X turn filling is 100% complete with 72/72
Answer: It will take 20.5 hours to completely fill this tank.
Analysis of measured filling process, from X = 42/72 share & from Y = 30/72 share
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length of the base=3 times height
b=3h;
h=b/3;
so, 1/2bh=24
1/2b(b/3)=24
(b^2)/6=24
b^2=144
b=12
21
40 km avrege speed
A = (B*H)/2
H = 2B
formula becomes A = (B*2*B)/2
this can be rewritten as A = (2*B^2)/2
the 2 in the numerator and denominator cancel out and you get:
A = B^2
(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3
a)0
b)206
c)0
d)250
e)39
f)92
perimeter = pi(r) + 2r
= 19.782 + (12.6)
= 32.382 cm
0,8x= y
y – 0,7y = 270
y= 270 / 0,3
y=900
0,8x=900
x=1125