This can be expressed as exponential growth of the form:
f=ir^t, f=final amount, i=initial amount, r=common ratio (rate), t=time
In this case we are given the point (2, 6500) and i=5000 so we can solve for the rate...
6500=5000r^2 divide both sides by 5000
1.3=r^2 take the square root of both sides and note that we know r>0
r=1.3^(1/2) so our equation becomes:
f=5000(1.3)^(1/2)^t and knowing that (a^b)^c=a^(b*c) we can say:
f=5000(1.3)^(t/2) so for t=18
f=5000(1.3)^9
f≈53022 (to nearest whole bacteria)
If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).
Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).
If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.
• Case 1: suppose <em>x</em> < 0. Then
<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = -1/√2 → <em>y</em> = ±1/√2
• Case 2: suppose <em>x</em> ≥ 0. Then
<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = 1/√2 → <em>y</em> = ±1/√2
In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.
Answer:
Assignment: 01.07 Laboratory TechniquesAssignment: 01.07 Laboratory Techniques
Step-by-step explanation:
Assignment: 01.07 LaboratoryAssignment: 01.07 Laboratory Techniques TechniquesAssignment: 01.07 Laboratory TechniquesAssignment: 01.07 Laboratory Techniques
8/ (1 1/2) = 36/x....8 muffins to 1 1/2 tsp = 36 ( 3 doz) muffins to x tsp
this is a proportion, so we cross multiply
(8)(x) = (1 1/2)(36)
8x = 3/2 * 36
8x = 108/2
8x = 54
x = 54/8
x = 6 3/4 <=== 6 3/4 tsps are needed to make 3 doz muffins
The answer could be something of the sort like 2/24, because 1/12 can't be simplified, you would just keep multiplying both numbers until you get another fraction equal to 1/12