first term.
a= 80
second term. a2=10
so Common ratio r=a2/a =10/80=1/8=0.125
checking for the validity of common ratio.
a2×r=10×1/8=10×0.125=1.25 =a3= Third term.
Hence it is true.
Now 4th Term =1.25/8=0.15625
It can be observed that 4TH term Has digit that is 5 places past the decimal point.
Hence, 0.15625 is the terminating number as per the question.
So the required sum is.
S=a+a2+a3+a4
->S=80+10+1.25+0.15625
->S= 91.40625
Hope it helps...
Regards,
Leukonov/Olegion.
Answer:
hey there hope this answer helps you out
Step-by-step explanation:
we have two functions f(x) and g(x) such that
and
solving for f ° g, we'll substitute the x in f(x) by the value of g(x)
taking x common in denominator
for a function to exist it should not have 0 in its denominator
checking the values of x for which the denominator of f ° g becomes 0 :-
so the function doesn't exist at values x = 0, -6
<h3>So, <u>0, -6</u> cannot be in the domain of f°g</h3>
Y=10.55
Z=17.53
Y: tan37=y/14 ; 14(tan37)=10.55
Z: sin53=14/z ; 14/sin53=17.53
Answer: −14−7
Step-by-step explanation:
(-8x + 2) - (6x +9)
-8x + 2 - 6x - 9
-14x - 7
