Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
Answer:
Translation (the +5)
Dilation (the x 4)
Compression (the 3 to 1/3, and the x to 2-x)
Step-by-step explanation:
Brainliest, please!
Answer:
a=1 or a=5/6
Step-by-step explanation:
I'm going to attempt to factor 6a^2-a-5
a=6
b=-1
c=-5
Find two numbers that multiply to be a*c and add to be b.
a*c=-30 =-6(5)
b=-1 =-6+5
So replace -a with -6a+5a in the expression we started with
6a^2-6a+5a-5
now we factor by grouping
6a(a-1)+5(a-1)
(a-1)(6a-5)
Now let's solve the equation:
(a-1)(6a-5)=0
So a=1 or a=5/6
Answer: Carla P. 15.1
Step-by-step explanation:
Answer:
8.85
Step-by-step explanation:
