The set of two-digit primes is {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
Of that list, the following primes are mirror images of each other
13 and 31
17 and 71
37 and 73
79 and 97
Note: we ignore 11 since 11 flips to 11 which is not distinct from its original
If you're looking for the largest prime of this form, then its 97
If you're looking for the largest gap, then subtract each pair
31-13 = 18
71-17 = 54
73-37 = 36
97-79 = 18
We see that 71 and 17 have the largest gap
Hi there
a/(a² - 36) + 2/(a-6) = 1/(a+6)
a/(a+6)(a-6) + 2/(a-6) = 1/(a+6)
Now we need to multiply the terms by (a+6)(a-6)
a + 2(a+6) = 1(a-6)
a + 2a + 12 = a - 6
3a + 12 = a - 6
3a - a = -6 - 12
2a = -18
Divide both sides by 2
2a/2 = -18/2
a = -9
Thus, The correct option is A
I hope that's help !
Good luck !
Answer:
There's one 0.
Step-by-step explanation:
Because I multiplied it and there's only one 0.
Answer:
Step-by-step explanation:
Solve
For x=-1/2 and y=-3
substitute the value of x and the value of y in the expression above
Remember that
so
Convert to mixed number
2 3/4 * 3.36 / (1 2/5)
=11/4 * 3.36 / (7/5)
= 9.24 / (7/5)
= 9.24 / 1.4
= 6.6
