Hello!
You do 72/6 which is 12.
So the answer is 12
Hope this helps!
We have that
csc ∅=13/12
sec ∅=-13/5
cot ∅=-5/12
we know that
csc ∅=1/sin ∅
sin ∅=1/ csc ∅------> sin ∅=12/13
sec ∅=-13/5
sec ∅=1/cos ∅
cos ∅=1/sec ∅------> cos ∅=-5/13
sin ∅ is positive and cos ∅ is negative
so
∅ belong to the II quadrant
therefore
<span>the coordinates of point (x,y) on the terminal ray of angle theta are
</span>x=-5
y=12
the answer ispoint (-5,12)
see the attached figure
Answer:
Below in bold.
Step-by-step explanation:
8 18 30 44
the differences are 10 12 14 .etc
1.1.1 Next 2 terms are 44+ 16 = 60 and 60+ 18= 78.
1.1.2 8 = 1^2 + 7
18 = 2^2 + 14
30 = 3^2 + 21
44 = 4^2 + 38
So the nth term = n^2 + 7n.
1.1.3 n^2 + 7n = 330
n^2 + 7n - 330 = 0
(n + 22)(n - 15) = 0
n = -22, 15
As the term sequence must be positive the answer is 15.
Answer:
14
Step-by-step explanation:
Ordinarily, a quick multiplication of 7 by other integers up to 10 indicates that only 7*9 yields 63, i.e ends with 3 as required.
Thus the set of possible multiples of the integer 7 ending with the digit 3 will form the arithmetic series with the first term being Ao = 63 and the common difference being d= 7*10= 70. That is we can see the series in details....
the nth term could be evaluated from the formular
An=Ao+(n-1)d (1)
The series could be explicitly depicted as follows:
9*7=63= 63+70*0
(10+9)*7=133 = 63+70*1
(20+9)*7=203=63+70*2
(30+9)*7=273=63+70*3
.................................
(130+9)*7=973=63+70*13
The last 'n' corresponding to the problem statement could be evaluated from equation (1), assuming An=1000:
1000=63+(n-1)*70
1000-63=70(n-1)
937/70=13.38=n-1
n=14.38
Thus the number of possible multiples of 7 less than 1000 ending with digit 3 will be 14.
Check: 7 times 142 is 994, so there are exactly 142 positive multiples of 7 less than 1000.
One tenth of these, ignoring the decimal fraction, end with a digit of 3.
Answer:
1/18 or 5.56% chance
Step-by-step explanation: