"Starting with three, every consecutive line has 2 less than twice the previous line."
this statement means that
your staring line has 3 marbles. You multiply the 3 marblesby 2 so
3x2=6
And then you minus it by 2
6-2=4
which means that you'll get 4 marbles for the next line.
So to get your 6th line, you count how many marbles is on the 5th line but since your diagram doesn't have the 5th line you have to figure out the 5th line by counting how.many marbles is on the 4th line.
4th line = 10 marbles
10×2=20
20-2=18
5th line = 18
18×2=36
36-2=34
So the 6th line has 34 marbles.
Answer:
Hence the function which has the smallest minimum is: h(x)
Step-by-step explanation:
We are given function f(x) as:
- f(x) = −4 sin(x − 0.5) + 11
We know that the minimum value attained by the sine function is -1 and the maximum value attained by sine function is 1.
so the function f(x) receives the minimum value when sine function attains the maximum value since the term of sine function is subtracted.
Hence, the minimum value of f(x) is: 11-4=7 ( when sine function is equal to 1)
- Also we are given a table of values for function h(x) as:
x y
−2 14
−1 9
0 6
1 5
2 6
3 9
4 14
Hence, the minimum value attained by h(x) is 5. ( when x=1)
- Also we are given function g(x) ; a quadratic function passing through (2,7),(3,6) and (4,7)
so, the equation will be:
Hence on putting these coordinates we will get:
a=1,b=3 and c=7.
Hence the function g(x) is given as:
So,the minimum value attained by g(x) could be seen from the graph is at the point (3,6).
Hence, the minimum value attained by g(x) is 6.
Hence the function which has the smallest minimum is h(x)
Answer:
Step-by-step explanation:
Correct option is
D
[1,(1+π)
2
]
f(x)=(1+sec
−1
(x))(1+cos
−1
(x))
Here the limiting component is cos
−1
(x), since the domain of cos
−1
(x) is [−1,1].
Therefore,
f(1)=(1+0)(1+0)
=1
f(−1)=(1+π)(1+π)
=(1+π)
2
Hence range of f(x)=[1,(1+π)
2
]
The midpoint would be (-2, -1)