All categories

3 years ago

Compare square inch for 8 inch square and 8 inch round circle

Mathematics

1 answer:

alexira [117]3 years ago

7 0

Answer:

An 8 inch Circle has a diameter of 8 inch, whereas the 8 inch square has length and width of 8 inch

Step-by-step explanation:

For a circle to be an 8 inch, the diameter should be 8 inches. Square's sides are equal, so we would have 8 inches on each side of the square. The area of this 8 inch square would be 64 in^2, whereas the 8 inch circle's area is 50.24 inch ^2. (because (3.14)(r^2))

You might be interested in

A model house has a scale of

Lapatulllka [165]

Answer: Assuming that you mean 1in:2ft the model house would be 13 inches wide.

Step-by-step explanation: well 1 is half of 2 right? So if you want to get the numbers easier to use then just divide that 26 by 2 and then you get the scale 1in:1ft. From there it is easy, 13 feet of the real house= 13 inches in the model house

5 0

1 year ago

How many lines of symmetry does a circle have? What point do all lines of symmetry for a given circle have in common?

Alina [70]

A circle has infinit lines of symmetry. They pass through the centre of the circle and are the circle's diameter. Hope this helps! (:

4 0

3 years ago

In a triangle the measurement of an exterior angle is always greater than the measurement of the adjacent interior angle.what is

Mariana [72]

The measure of and adjacent angle is less than the measure of the exterior angle

7 0

3 years ago

Cartos saves aluminum cans to sell to the recycling center. He recently recycled 43 pounds of cars and received $23.65. How many

aniked [119]

Uh add and than subtract you will get the answer jsjskskskskskeke.

6 0

3 years ago

Integrate 1 - x / x(x2 + 1) d x by partial fractions.

solniwko [45]

Answer:

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

Step-by-step explanation:

step 1:- by using partial fractions

$[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }$ ......(1)

step 2:-

solving on both sides

$1-x=A(x^{2} +1)+(Bx+C)x......(2)$

substitute x =0 value in equation (2)

1=A(1)+0

A=1

comparing x^2 co-efficient on both sides (in equation 2)

0 = A+B

0 = 1+B

B=-1

comparing x co-efficient on both sides (in equation 2)

-1 = C

step 3:-

substitute A,B,C values in equation (1)

now

$\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 } d x -\int\limits \frac{1}{x^{2}+1 } d x$

by using integration formulas

i) by using $\int\limits \frac{1}{x} d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c\\$ .....(b)

$\int\limits tan^{-1}x dx =\frac{1}{1+x^{2} } +C$ .....(c)

step 4:-

by using above integration formulas (a,b,and c)

we get answer is

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

6 0

3 years ago