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Taking this example into account, we can see that setting the first value equal to 1, we obtain that 
and 
. Using this information, we find that 
. It shows that when x is positive, the succussive terms are increasing.
Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that
and x=-40. In the next iteration, 
. In the next iteration, 
. By this way, we find that even if the initial value is less than zero, value of the successive iterations is increasing.
Using the function
and taking the initial value equal to 4, we find that 
and x=-2. In the next iteration, 
. If we continue the iterations we'll see that they are decreasing.
Setting the initial value equal to 2, we find that
and x=0. The next iteration is 
. In this case, the interations are also decreasing.
If we set the initial value equal to 1, we find that
and x=1. In the next iteration, 
and the iterations are decreasing.
Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that
Using the function
Setting the initial value equal to 2, we find that
If we set the initial value equal to 1, we find that
Answer:
1. 121 π unit²
2. 143°
3. 151 unit²
Step-by-step explanation:
1.
Area of a circle is given by the formula A = πr²
where
A is the area,
r is the radius of the circle
From the given diagram, we can see that the radius is 11, hence the area will be:
The answer is units^2
2.
The unshaded secctor and the shaded sector equals the circle. We know that circle is 360°. The unshaded sector has an angle of 217°. So the shaded part will be 360 - 217 = 143°
The measure of the central angle of the shaded sector is 143°
3.
Area of a sector is given by the formula
Where
is the central angle of the sector (in our case it is 143°)
r is the radius (which is 11)
Plugging in all the info into the formula we have:
<em>rounding to the nearest whole number, it is </em>151 units^2
The zeros of given function is – 5 and – 3
<u>Solution:</u>
We have to find the zeros of the function by rewriting the function in intercept form.
By using intercept form, we can put value of y as to obtain zeros of function
We know that, intercept form of above equation is
Taking “x” as common from first two terms and “3” as common from last two terms
x (x + 5) + 3(x + 5) = 0
(x + 5)(x + 3) = 0
Equating to 0 we get,
x + 5 = 0 or x + 3 = 0
x = - 5 or – 3
Hence, the zeroes of the given function are – 5 and – 3