Answer:
16,17 and 18
Step-by-step explanation:
In statistics mode of a set of entries is the entry which is repeated maximum. There can be more than one mode in a set of entries. These are called mode,
Bimode ( two mode ) , trimode ( three mode ) and Multimode ( four or more ) .
Hence here our set of entries is,
20,17,16,17,18,16,18,19
arranging them in ascending order
16,16,17,17,18,18,19,20
hence in this case we see that 16,17and 18 all are getting repeated for two times, the maximum.
Hence we have a trimode here
16,17,18
Answer:
45
Step-by-step explanation:
Answer:
∠B and ∠F
∠F and ∠E . . . already listed in the problem statement, so may not be considered an "other pair"
Step-by-step explanation:
∠C and ∠F are vertical angles, so are congruent. Then any angle complementary to one of those will also be complementary to the other.
Likewise, ∠B and ∠E are vertical angles and congruent. Any angle complementary to one of them will also be complementary to the other. Here, ∠E and ∠F are listed as complementary, so we know ∠B and ∠F will be also.
3x^2 – 5x – 12 = (3x+4<span>)</span>(x-3)
x^2 -9 = (x-3<span>)</span>^2
<span>8x^3+ 12x^2 - 2x - 3 = (4x^2 -
1)(2x + 3)
</span>12x^2 +16x -3= (6x - 1)(2x + 3)
x^2 + 8x - 9= (x + 9<span>)</span>(x - 1) (I <span>can not</span> see
who it can all be positive<span> ;</span>_;<span> )</span>
<span>3(x + y) + a(x + y)= (3x + 3y) + (ax + ay)</span>
