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:
it a
Step-by-step explanation:
hi
Answer:
234.85
Step-by-step explanation:
This was easy!!!
(The gardener comes 4 times a month... so divide 85.40 by 4 and you will get 21.35. Each time he comes, he gets 21.35 and by the end of the month, he has 85.40. It says he already came 11 times. So 4 + 4 = 8 + another 4 that equals 12. He came only two months and a few weeks. This means multiply 85.40 by 2 so you would get 170.8. It said he came 11 times so substract 11 from 8 and you would get 3. So multiply 21.35 by 3 and get a total of 64.05. Add 170.8 and 64.05 to get 234.85!!! The answer is 234.85 or C!!!!
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