The area of the shaded region is 40π/3 square cm if the radius of the small circle r is 3 cm and the radius of the large circle R is 7 cm.
<h3>What is a circle?</h3>
It is described as a set of points, where each point is at the same distance from a fixed point (called the center of a circle)
We have a circle in which the shaded region is shown.
The radius of the small circle r = 3 cm
The radius of the large circle R = 3+4 = 7 cm
The area of the shaded region:
= area of the large circle sector - an area of the small circle sector
= (120/360)[π7²] - (120/360)[π3²]
= 49π/3 - 3π
= 40π/3 square cm or
= 13.34π square cm
Thus, the area of the shaded region is 40π/3 square cm if the radius of the small circle r is 3 cm and the radius of the large circle R is 7 cm.
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Answer:
A and D both have a value of -4
Step-by-step explanation:
A. (-4) = -4
B. |-4| = positive value of what is inside = 4
C. |4|
= positive value of what is inside = 4
D. -|4| = -positive value of what is inside = -4
Answer:
Hey there!
We can write this equation, where x is the hourly rate and y is the fixed fee.
4x+y=270
7x+y=420
4x+y=270
-7x-y=-420
-3x=-150
x=50
4(50)+y=270
200+y=270
y=70.
The fixed fee is G. $70.
Let me know if this helps :)
Using the z-distribution, as we have the standard deviation for the population, it is found that the smallest sample size required to obtain the desired margin of error is of 77.
<h3>What is a z-distribution confidence interval?</h3>
The confidence interval is:

In which:
is the sample mean.
is the standard deviation for the population.
The margin of error is given by:

In this problem, we have that the parameters are given as follows:
.
Hence, solving for n, we find the sample size.






Rounding up, the smallest sample size required to obtain the desired margin of error is of 77.
More can be learned about the z-distribution at brainly.com/question/25890103
1.
has a horizontal asymptote at 
This means that

(for at least one of these limits)
2.
has a vertical asymptote at 
This means that
has a non-removable discontinuity at
. Since
is some rational function, there must be a factor of
in its denominator.
3.
has an
-intercept at (1, 0)
This means
.
(a) With

the second point above suggests
. The first point tells us that

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if
so that the linear terms vanish.
The third point tells us that

So

(b) Since

we find that
, and
and
.