8.6667
91.1154
No. A factor of a number divides it evenly.When you try to divide 26 by 3, you get a remainder of 2. That means 3 is not a factor.
26 and divide 5278 = 0.0049261083743842365
0.0663
You do the opposite of what the question is asking you to do. So if the question is asking you to add, you subtract instead. example: 26a =78 26a=78 (26*a =78) _______ 26 (divide by 26) a=3 hope this helped.<3
91.1154
3/26 is in its most simplified form . Reason , there are no common factors between '3' and '26'.
No. A factor of a number divides it evenly.When you try to divide 26 by 3, you get a remainder of 2. That means 3 is not a factor.
26 and divide 5278 = 0.0049261083743842365
Typically, it is 4:3. As an example: 26" across the screen and 19.5" down. If you divide 19.5 by 3 you get 6.5; then, if you divide 26 by 6.5, you get 4. So, dividing both 26 and 19.5 by 6.5, you get 4 and 3; hence, the 4:3 designation (nothing magic about 6.5--it is simply the divisor which reduces the 26 and 19.5 to lowest possible integers.)
31 + 26 + 24 = 81 / 3 = 27 To find the above average, add the the three numbers together, then divide by 3
0.1154
To determine how many times 3 will go into 26, we perform division. When we divide 26 by 3, we get 8 with a remainder of 2. Therefore, 3 will go into 26 eight times evenly, with a remainder of 2.
0.0385
To divide 7903 by 26, you would perform long division. Start by dividing 79 by 26, which equals 3 with a remainder of 1. Bring down the next digit, 0, to the remainder to make it 10. Then divide 10 by 26, which equals 0 with a remainder of 10. Finally, bring down the last digit, 3, to the remainder to make it 103. Dividing 103 by 26 gives you 3 with a remainder of 25. Therefore, 7903 divided by 26 equals 303 with a remainder of 25.
8.6667
The proof is by the method of reductio ad absurdum. We start by assuming that cuberoot of 26, cbrt(26), is rational. That means that the cube root can be expressed in the form p/q where p and q are co-prime integers. That is, cbrt(26) = p/q.Therefore, p^3/q^3 = 26 which can also be expressed as 26*q^3 = p^3 Now 26 = 2*13 so 2 divides the left hand side (LHS) and therefore it must divide the right hand side (RHS). That is, 2 must divide p^3 and since 2 is a prime, 2 must divide p. That is p = 2*r for some integer r. Then substituting for p gives, 26*q^3 = (2*r)^3 = 8*r^3 Dividing both sides by 2 gives 13*q^3 = 4*r^3. But now 2 divides the RHS so it must divide the LHS. That is, 2 must divide q^3 and since 2 is a prime, 2 must divide q. But then we have 2 dividing p as well as q which contradicts the requirement that p and q are co-prime. The contradiction implies that cbrt(26) cannot be rational.