Answer:
12
Step-by-step explanation:
Given that
Current count of pushups = 8
Number of increase in pushups each day = 2
The desired pushup goal on daily basis = 30
Days require = x= ?
as we can conclude the number of pushups would increase by following sequence
8, 10, 12, 14, 16, ......30
Now as we can be the number of pushups by a constant number each day it means this sequence is arithmetic.
First term of our sequence is X1 and the final term is Xn, and the difference between two consecutive terms (Common Difference) is 2.
So
X1 = 8
Xn = 30
d = 2
Using the formula of nth term for arithmetic sequence
Xn = X1 + (n-1)d
Substituting the values
30 = 8 + (n-1)2
30-8 = (n-1)2
22 = 2n -2
22+2 = 2n
2n = 24
n = 24/2
n = 12
So, it means on the 12th day he will have to do 30 pushups if he starts from 8 and increase 2 pushups each day.
If you want to round to the nearest thousand, you cannot look at the thousands place to round. You will have to look at the number in the hundreds place. The rules of rounding are
<em>5 and above-round up. Example-2743. 7 is greater than 5, so 2743 rounded to the nearest thousand is 3000.
</em>
<em>4 and below-round down. Example-4390. Three is less than 5, so 4390 rounded to the nearest thousand is 4000.
</em>
Apply these rules to your problem. 8 is greater than 5. So we will round up. You get 5000.<em>
</em>
Answer:
8 cups of shampoo concentrate
Step-by-step explanation:
let x=how much shampoo concentrate Devin should add
cross multiply
2*36=9x
72=9x
72/9=x
x=8
Devin should add 8 cups of shampoo concentrate to the water to make the shampoo solution.
If you are having trouble with ratios, turn them into fractions.
ex; 3 : 5 = 3/5, 8 : 10 = 8/10, 6 : 15 = 6/15
After you have turned all of your ratios into fractions, you can find a common denominator for all the fractions.
ex; 3/5, 8/10, 6/15 ⇒ 3/5, 4/5, 2/5
Now it you can easily order the ratios from least to greatest.
ex; 6 : 15, 3 : 5, 8 : 10
Yes it is possible for a geometric sequence to not outgrow an arithmetic one, but only if the common ratio r is restricted by this inequality: 0 < r < 1
Consider the arithmetic sequence an = 9 + 2(n-1). We start at 9 and increment (or increase) by 2 each time. This goes on forever to generate the successive terms.
In the geometric sequence an = 4*(0.5)^(n-1), we start at 4 and multiply each term by 0.5, so the next term would be 2, then after that would be 1, etc. This sequence steadily gets closer to 0 but never actually gets there. We can say that this is a strictly decreasing sequence.
If your teacher insists that the geometric sequence must be strictly increasing, then at some point the geometric sequence will overtake the arithmetic one. This is due to the nature that exponential growth functions grow faster compared to linear functions with positive slope.
