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You could measure the circumference and the diameter of a circle and divide the circumference by the diameter to get an estimate for pi.

thank you :)

If  reduced by 15%,  the cost of the item must just be 85%  of what it originally was...

85% of 14.95  =

.85 * 14.95   ≈

 $12.71 

1/4 + 1/2   =

1/4  +  2/4  =

3/4

LOL!!!

Yep.....It would have helped to supply the link  :

https://www.desmos.com/calculator/2thymd2pwe

LOL !!!

Looks like the shaded regions are also absent today 

We can solve this system....where a1 is the first term and d is the common difference between terms

30   =  a1 + 6d         

60  =  a1  + 10d

Subtract the 1st equation from the 2nd

30  =  4d       divide both sides by 4

7.5  =  d

And using the second equation to find a1, we have

60  =  a1  + 10(7.5)

60  = a1  +  75           subtract 75 from both sides

-15    =  a1

So......the 21st term is

-15 + 7.5(20)  =

135

Don't see any solution presented, but.....it will lie in the over-lapping shaded regions shown here :

2^x   =   13                     Take the log of both sides.

log 2^x   =   log 13

x log 2   =   log 13        Divide both sides by log 2 .

x   =   log 13 / log 2      Put this in a calculator to get that

x   ≈   3.700

In an arithmetic sequence, the 7th term is 30, and the 11th term is 60. What is the 21st term of this sequence?

11 - 7 + 1 =5 number of terms

60 - 30 =30 - the difference between above 5 terms.

30 / 5 = 6 - the common difference. The nth. term is given by:

F + (N - 1) x D =nth. term, where F=first term, N=number of terms, D=common difference.

First term =30 - (6*6) = - 6

-6 + (21 -1) x 6 =

-6 + 20 x 6 =

-6 + 120 = 114 - the 21st. term.

P.S. There is something wrong with your sequence as stated. Either the 7th term or 11th term is wrong!. If the 7th term is right, then the 11th term is 54 and the 12th term is 60. If the 11th term is 60, then the 7th term should be 36. Either way, if the 1st term is 0, then the 7th should be 36 and the 11th should be 60.

If the 1st term is -6, then the 7th term should be 30 and the 11th term should be 54.

Check your sequence carefully.

I tested the method on another number:

\(0.\overline3=x \\~\\ 3.\overline3=10x \\~\\ 3.\overline3-0.\overline3=10x-0.\overline3 \\~\\ 3.\overline3-0.\overline3=10x-x \\~\\ 3=9x \\~\\ \frac{1}{3}=x\)

It worked! .....Now I will try it on \( \overline9 \) .

\(\overline9=x\)

But what is \( \overline9 \) times 10 ?  I need to move the decimal point to the right one place, or add a zero at the "end." But I can't reach the end... it's too far away !   And, if I just try to avoid the issue by writing this...

\(10\,\cdot\,\overline9=10x \\~\\ 10\,\cdot\,\overline9-\overline9=10x-x \\~\\9\cdot\overline9=9x \\~\\ \overline9=x\qquad\text{I get exactly what I started with!}\)

Thanks a lot!!!

Solve for x:
4 - (x)/(4) = 6

Put each term in 4 - (x)/(4) over the common denominator 4: 4 - (x)/(4) = 16/4 - (x)/4:
16/4 - x/4 = 6

16/4 - x/4 = (16 - x)/4:
(16 - x)/4 = 6

Multiply both sides of (16 - x)/4 = 6 by 4:
(4 (16 - x))/4 = 4×6

(4 (16 - x))/4 = 4/4×(16 - x) = 16 - x:
16 - x = 4×6

4×6 = 24:
16 - x = 24

Subtract 16 from both sides:
(16 - 16) - x = 24 - 16

16 - 16 = 0:
-x = 24 - 16

24 - 16 = 8:
-x = 8

Multiply both sides of -x = 8 by -1:
(-x)/(-1) = -8

(-1)/(-1) = 1:
Answer: | x = -8

2 = 3/4 + x

2 -3/4 = x 

2 - .75 =x

 1.25 =x

First, divide 27000/ 2, then multiply that result by 1  to get 13500

It's just the area of a circle times the height of the cylinder, to find the volume of the cylinder.

Next time, use the web2.0 calculator.

From ,wikihow. ... Divide the new number by your divisor, and write the result above the dividend as the next digit of the quotient. In the example, determine out how many times 6 can go into 10. Write that number (1) into the quotient above the dividend. Then multiply 6 by 1, and subtract the result from 10.

80/100 * 50.... 50 becomes 1, and 100 becomes 2. 80/ 40 =2, 

Guest, there's 1 basic rule. Anything to the power of 0, is one. So, \(2^0\)=1

Very Nice way, Hecticar!

80 = 2^4 * 5   [These are called PRIME factors of 80]

80 =1 | 2 | 4 | 5 | 8 | 10 | 16 | 20 | 40 | 80 [These are called DIVISORS of 80].

Two squares have a combined area of 68. The side of one square is 4 times the length of the side of the other square. Find the perimeter of each square.

Let the side of the 1st. square =S

Then the side of the 2nd. square =4S

Area of 1st. square =S^2

Area of 2nd. square =(4S)^2 =16S^2. But we have:

S^2 + 16S^2 = 68

17S^2 = 68    divide both sides by 17

S^2 =4 take the square root of both sides

S =2 - the side of the 1st. square.

4 x 2 =8 - the perimeter of the 1st. square = the side of the 2nd. square.

8 x 4  =32 - the perimeter of the 2nd. square.

or 3.7L