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:
a. 60/61
Step-by-step explanation:
With reference angle A
base(b) = 60
hypotenuse (h) = 61
Now
Cos(A) = b / h
= 60/ 61
Check the picture below.
the segment LOJ is a diameter, and therefore the arcLCJ made by that diameter will have a central angle of 180°.
that simply means that the arcLC is 180 - 94, or 86°.
the ∡LJC is an inscribed angle that is intercepting the arcLC, and therefore, by the inscribed angle theorem, it'll be half of arcLC.
15. the answer is 4/10 I think