6 3
------
250
ignore this sentence
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:
9^8
Step-by-step explanation:
The total number of ants is
(total number of sections)(ants in each section).
let x be the total number of ants.
Substitute the given values.
x = (9^2)(9^6)
x = 9^(2+6) <= exponent rule: when multiplying exponents with the same base, add the exponents.
x = 9^8
First we find k by using the two values given:
P=Ae^(kt), 8 years between-->t=8
199 = 195•e^(8k)
Log 199 = Log [195•e^(8k)]
Log 199 = Log 195 + 8k
8k = Log 199 - Log 195
k = 0.00882/8 = .0011
Next, we plug the new data in using this k:
14 years between 2002-2016-->t=14
P = Ae^my
P = 199mi.•e^(14•.0011)
P = 199mi.•e^(.0154)
P = 199mi. • 1.0155 = 202,088,000
Answer:
cest trop facile mes mois jai besoin d'aide alor quesque quil faut faire di mois
