A: 1/8b = 2 1/4
B: 2 1/4 / 1/8 = 18
Ryan filled 18 bags.
When working with very large or very small numbers, scientists, mathematicians, and engineers use scientific notation to express these quantities. Scientific notation is a mathematical abbreviation, based on the idea that it is easier to read an exponent than to count many zeros in a number. Very large or very small numbers need less space when written in scientific notation because the position values are expressed as powers of 10. Calculations with long numbers are easier to do when using scientific notation
I'm pretty sure it depends what kind of triangle it is
Answer:
x = 30
Step-by-step explanation:
here 50 is hypotenuse as it is opposite of 90 degree.
x and x + 10 are the two other smaller sides of a right angled triangle respectively.
using pythagoras theorem,
a^2 + b^2 = c^2
x^2 + (x + 10)^2 = 50^2
x^2 + x^2 + 20x + 100 = 2500
2x^2 + 20x + 100 = 2500
2x^2 + 20x + 100 - 2500 = 0
2x^2 + 20x - 2400 = 0
2(x^2 + 10x - 1200) = 0
x^2 + 10x - 1200 = 0
x^2 + (40 - 30) - 1200 = 0
x^2 + 40x - 30x - 1200 = 0
x(x + 40) - 30(x + 40x) = 0
(x + 40)(x - 30) = 0
either x + 40 = 0 OR x - 30 = 0
x = 0 - 40
x = -40
x - 30 = 0
x = 30
x = -40,30
since the length and distance is not measured in negative ur answer will be 30
credit goes to sreedevi102
thank u very much . At first i was wrong and giannathecookie i m really sorry
The area formula of a circle is 
× π.
With that being said, we can solve both problems!
Starting with part A:
-The radius we know is 21cm. From there, we can substitute in our formula with the new value.
x π
From there, we can square our radius:
(21 x 21) x π
(441) x π
And finally, we can multiply by pi to get our answer.
(441) x π
1385.44236023
Which rounded to the nearest tenth is 1385.4. Therefore, the area of the porthole is 1385.4 centimetres.
The second part can be followed with the same process, hopefully you understand the concept enough to where you're comfortable solving problems like this on your own. Feel free to reach out if you have any further trouble!
<em>Hope this helped! :)</em>
