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Hunter-Best [27]
2 years ago
11

Red Hook has a population of 6,200 people and is growing at a rate of 8% per year. Rhinebeck has a population of 8,750 and is gr

owing at a rate of 6% per year. In how many years, to the nearest year, will Red Hook have a greater population than Rhinebeck? Show the equation or inequality you are solving and solve it graphically.​
Mathematics
1 answer:
Kruka [31]2 years ago
7 0

Answer:

Step-by-step explanation:

subtract 8750 to 6200

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Need help with this question please!
nevsk [136]
I would say it is A because if you subtract <em>p,</em> the original price by $2.50, you would get <em>d, </em>the discounted price. Look at B u see that you're adding the discount which doesn't make sense. Looking at C, the discounted price of different prices can't always be the same. And finally, D, the discounted price is greater than the original. Also, if you subtract you would get different discounts.
6 0
3 years ago
Solve for -5v4(-31)​
Ilia_Sergeevich [38]

Answer:

155v^4

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
1. 10(.15)=1.5
Fed [463]

Answer:

6.14125(0.15) = 0.9211875 (below 1)

6.14125 - 0.92118 = 5.22007

Step-by-step explanation:

Given data

1. 10(.15)=1.5

2. 10-1.5=8.5

3. 8.5 (.15)=1.275

4. 8.5-1.275=7.225

continuation the sequence

5)    7.225 (0.15) = 1.08375

6)    7.225 -  1.08375 = 6.14125

7)  <u> 6.14125(0.15) = 0.9211875   (below -one)</u>

8 ) <u> 6.14125 - 0.9211875 = 5.2200625 (get number 5)</u>

9)   5.2200625(0.15) = 0.783009

10) 5.2200625 - 0.783009 = 4.4370532

3 0
2 years ago
Can somebody prove this mathmatical induction?
Flauer [41]

Answer:

See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

So, you have proved the initial statement

4 0
3 years ago
If i have 18 apples and i get 1328 more how many apples would i have
devlian [24]

Darling, this problem is quite simply addition. You add 18 to 1328. 1328 + 18. we add the eights first. 8 + 8 equals 16. The 6 stays and we carry the one. 1 + 2 is 3. And then we add our carried 1 to make 4. So, the answer is 1346. I hope this helped!

6 0
3 years ago
Read 2 more answers
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