Answer: There is 22 possible combinations.
Step-by-step explanation:
%a is equal or greater then 11 and it is odd
count=0;
P=[];
for a=11:2:499
b=a+2;
%assume a and b are prime;
prime=1;
%check if a and b are prime
for k=2:ceil(sqrt(b))
%if a OR b is divisible by k, break
%it is not prime
if mod(a, k)==0 | mod(b, k)==0
prime=0;
break;
end
end
if prime
count=count+1;
P(count, :)=[a b];
end
end
fprintf('There is %d possible combinations.\n', count)
display(P)
There is 22 possible combinations.
P =
11 13
17 19
29 31
41 43
59 61
71 73
101 103
107 109
137 139
149 151
179 181
191 193
197 199
227 229
239 241
269 271
281 283
311 313
347 349
419 421
431 433
461 463
Answer:
x = 8
Step-by-step explanation:
4x + 112 = 9x + 72
-112 -112
4x = 9x - 40
-9x -9x
-5x = -40
----- --------
-5 -5
x = 8
Remark
Good thing you tried to answer the question. At lease we know how to proceed.
A is wrong, because you can draw a conclusion.
B Remember that CD is constructed so that AB is cut in half. So the distance from CD to B is the same distance as CD to A. Choice B is true.
C is incorrect. CD can be extended to any length as long as CD goes through the point where the two arcs above and below AB meet are joined by C and D, then CD can be any length. In order words, CD can be any length.
D Doesn't have to be. See the answer for C.