245 r = 100 % 24.5 r = 10 % 49 r = 20 %
A/5 can only be a mixed number if A is greater than 5. But since we don't know what A is, we can't answer that specifically.
It is 22 + R
All multiples of 6 are even numbers.
The number 41 is a prime number. A prime number has only 2 factors which are 1 and itself. Composite numbers are everything else except 1 and 0. 1 and 0 are neither prime, nor composite.To test a number to find out if it is prime or composite, try dividing it by the first few prime numbers:41 ÷ 2 = 20 r 141 ÷ 3 = 13 r 241 ÷ 5 = 8 r 141 ÷ 7 = 5 r 6We don't have to test 4 or 6, since they are both divisible by 2. The divisor (7) is now larger than the result (5 remainder 6), so we can stop. 41 is prime.
It depends on the value of r, which you have not specified.
how do you create a decimal or a mixed number that is either greater or less than any number
t < r
Yes. You have to have at least a 609 certification, and you can only buy in containers less than 20# with that certification. Look for on line 609 test on a Google search. Anything more than 20# needs a different certification.
you r asking a number whose twice will be 4 but less than 8. so, take a no. x. by ques. 2x=4; 2x=4; x=4/2; hence x=2; 2 is answer and less than 8 of twice it.
If the atomic number of Element D is 20, then Element D is calcium (Ca). The atomic number of calcium is 20, so R would also have an atomic number of 20. This means that R would also be calcium (Ca).
R-134a is 95% less damaging to the ozone layer than R-12.........
A number which is not a whole number. It could be a fraction whose absolute value is less than one or a mixed (r improper) fraction, an irrational number, an imaginary or complex number, and so on.
R less than 0.3
Yes, as long as R is less than (4/5) ■
'r' is equal to, or anywhere between, -13 and 38 .
To calculate the number of 3-number combinations from the numbers 1-20, we can use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n = 20 and r = 3. Plugging these values into the formula, we get 20C3 = 20! / 3!(20-3)! = 1140. Therefore, there are 1140 possible 3-number combinations from the numbers 1-20.