Substitute the value of the variable into the equation and simplify.
=8+(3)(6)
=8+18
=26
Hope it helps!
The N6 certificate is on NQF level 5, and the Diploma is on NQF level 6.
It is n
the grove n6
n6
n times 6
Yes, here's the proof. Let's start out with the basic inequality 81 < 83 < 100. Now, we'll take the square root of this inequality: 9 < √83 < 10. If you subtract all numbers by 9, you get: 0 < √83 - 9 < 1. If √83 is rational, then it can be expressed as a fraction of two integers, m/n. This next part is the only remotely tricky part of this proof, so pay attention. We're going to assume that m/n is in its most reduced form; i.e., that the value for n is the smallest it can be and still be able to represent √83. Therefore, √83n must be an integer, and n must be the smallest multiple of √83 to make this true. If you don't understand this part, read it again, because this is the heart of the proof. Now, we're going to multiply √83n by (√83 - 9). This gives 83n - 9√83n. Well, 83n is an integer, and, as we explained above, √83n is also an integer, so 9√83n is an integer too; therefore, 83n - 9√83n is an integer as well. We're going to rearrange this expression to (√83n - 9n)√83 and then set the term (√83n - 9n) equal to p, for simplicity. This gives us the expression √83p, which is equal to 83n - 9√83n, and is an integer. Remember, from above, that 0 < √83 - 9 < 1. If we multiply this inequality by n, we get 0 < √83n - 9n < n, or, from what we defined above, 0 < p < n. This means that p < n and thus √83p < √83n. We've already determined that both √83p and √83n are integers, but recall that we said n was the smallest multiple of √83 to yield an integer value. Thus, √83p < √83n is a contradiction; therefore √83 can't be rational and so must be irrational. Q.E.D.
160ml
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3n + 14 = 83n = 8 - 14 = -63n/3 = -6/3n = -2The value of n is -2.
40 minutes drive or less now that N6 has been completed
275 kilometres (171 miles) taking this route:Take N62 ROSCREA, from Thurles, to N6 GALWAY (to N61 ROSCOMMON).Take N6 on the Athlone Bypass across to N61 ROSCOMMON at J2.Take N61 to N4 SLIGO, outside of Boyle.Take N4 to N15 LIFFORD, at Sligo.Take N15 to DONEGAL.
Try this link: http://www.nature.com/ng/journal/v37/n6/abs/ng1553.html