3.572
Area of a circle in terms of radius:
Area = π · r2 = 3.14 · 1.072 = 3.57 square meters.
In terms of diameter:
Area = π · (d
2
)2 = 3.14 · (2.13
2
)2 = 3.14 · (1.07)2 = 3.57 square meters.
In terms of circumference:
Area = C2
4π
= 6.72
4π
= 44.89
(4 · 3.14)
= 44.89
12.56
= 3.57 square meters.
Note: for simplicity, some results may be rounded to nearest hundredth and π was rounded to 3.14. See formula details below in this page.
A circle of radius = 1.066 or diameter = 2.133 or circunference = 6.7 meters has an area of 3.572 square meters which is equal to:
3.572E-6 square kilometers (km²)
35720 square centimeters (cm²)
0.0008827 acres (ac)
0.0003572 hectares (ha)
1.379E-6 square miles (mi²)
4.272 square yards (yd²)
38.45 square feet (ft²)
5537 square inches (in²)
Formulae:
Circle area formula
DONT CLICK ON THE LINKS! It’s 8.750 BY THE WAY the radius is half the diameter so just add the radius by itself or multiply by 2
Answer:
right
Step-by-step explanation:
The side ratios are ...
24 : 45 : 51 = 8 : 15 : 17
These numbers are a Pythagorean triple. The triangle is a right triangle.
__
24² + 45² = 576 +2025 = 2601 = 51² . . . . the Pythagorean theorem is satisfied
Answer:
<em> 50° </em>
Step-by-step explanation:
x° = 180° - (102° + 28°) =<em> 50°</em>
Answer:

Step-by-step explanation:
So we have the indefinite integral:

This is the same thing as:

So, let's do integration by parts.
Let u be (ln(x))². And let dv be (1)dx. Therefore:

Simplify:

And:

Therefore:

The x cancel:

Move the 2 to the front:

(I'm not exactly sure how you got what you got. Perhaps you differentiated incorrectly?)
Now, let's use integrations by parts again for the integral. Similarly, let's put a 1 in front:

Let u be ln(x) and let dv be (1)dx. Thus:

And:

So:

Simplify the integral:

Evaluate:

Now, we just have to simplify :)
Distribute the -2:

And if preferred, we can factor out a x:

And, of course, don't forget about the constant of integration!

And we are done :)