Answer:
7.
Solution given;
male=15
female=27
1st term=5*3
2nd term=3*3*3
now
Highest common factor=3
So
<u>The</u><u> </u><u>maximum</u><u> </u><u>number</u><u> </u><u>of</u><u> </u><u>groups</u><u> </u><u>that</u><u> </u><u>the</u><u> </u><u>teacher</u><u> </u><u>can</u><u> </u><u>make</u><u> </u><u>is</u><u> </u><u>3</u><u>.</u>
<u>and</u><u> </u><u>each</u><u> </u><u>team</u><u> </u><u>contains</u><u> </u><u>5</u><u> </u><u>male and</u><u> </u><u>9</u><u> </u><u>female</u><u>.</u>
Answer:
<h3>
B.</h3>
Step-by-step explanation:
The distance between two any points <em>a</em> i <em>b</em> is |a-b| as we don't know which of them is larger number.
So:
a=-2 and b=1 means distance |-2-1| = |-3| = 3
{If we know the value of given points we can subtract the smaller one from the larger. It also works: -2<1, so the distance would be 2-(-1)=2+1=3}
I do believe it would be x=30