Answer:
r=41/2
Step-by-step explanation:
You have to divide the diameter, 41, by 2 because the radius of the circle is half of the diameter. So, the answer would be 20.5
We know that
circumference=pi*D
where
D is the diameter
D=91 cm
so
pi-----> is the value of the calculator------> 3.1415926535897932
circumference=pi*91------> 285.88 cm------> 285.9 cm
the answer is
285.9 cm
Answer:
It should be 1,628
.
Step-by-step explanation:
We see two shapes in the figure and those shapes are a rectangle and a circle. Let's find the area of the circle first. 25m won't help us find the area of the circle so let's pretend that 25m isn't there for now. 40m seems to be the diameter of the circle and to find the area of the circle we need to multiply the radius squared by Pi or 3.14. Half of 40 is 20 so we can multiply 20 squared by Pi to give us 1,256.63706. Do not round this number yet. As you can see this circle isn't a full circle. It's a semicircle. We can divide 1,256.63706 by 2 to find the area of a semicircle. You should get 628.31853 and round that to the nearest tenth and we get 628.32. Now let's keep that number in mind and find the area of the rectangle. Using a calculator we can easily multiply 40 and 25 to find the area of the rectangle, which is 1,000. Add 1,000 and 628.32 and our final results should be 1,628.32. I don't see this number in your options but option B is closely related to this answer. I hope this helps and if this answer is wrong then please give me some feedback on what I did wrong! Thank you!
2. x = 1/4 ( 0.25)
4. y = 122, x = 42
6. x = 63, y = 8
8. is missing
10. 13.5 cm
12. x = 10.6
14. x = 5.45
Answer:
(2)
Step-by-step explanation:
Our logarithmic expression is:
.
Remember the logarithmic property that ln(a/b) = lna - lnb. So, we can write this as:

Also, we can write square roots as powers of one-half, so √e =
. There's another log property that:
. We can apply that here for both the √e and the y³:

Finally, note that ln(e) is just 1, so we have:

The answer is thus (2).
<em>~ an aesthetics lover</em>