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DochEvi [55]
3 years ago
10

What’s common difference 35,32,29,26

Mathematics
1 answer:
Vlad1618 [11]3 years ago
7 0
3 is the common difference
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Which of the following describes the following graph?
pychu [463]

<u>Answer:</u>

The correct answer option is B. both a function and a relation.

<u>Step-by-step explanation:</u>

We are given a from from which we can see that for each input we have exactly one output.

This means that we have a function because each element in the domain is matched with exactly one element in the range.

It is also a relation since each input related to the out put in some way.

Therefore, the correct answer option is B. both a function and a relation,

8 0
3 years ago
your kitchen is in your house is being remolded. it takes 2 workers , 12 days to finish the job. Each worker works 7 hours a day
uranmaximum [27]
I think it is $4284 b/c you  would do $25.5*7 hours*12 days*2 workers=$4284 total
3 0
3 years ago
Helpppp please thanksssss .
Elza [17]

Answer:

I think the answer is B

Step-by-step explanation:

3 0
3 years ago
What is the factorization of the polynomial below<br>-x^2 - 15x - 56​
Andrej [43]

Answer:

Step-by-step explanation:

This is more easily understood if you take out a common factor of -1

-1(x^2 + 15x + 56)

7*8 = 56

7 + 8 = 56

-1 (x +8)(x+7)

3 0
2 years ago
For the years from 2002 and projected to 2024, the national health care expenditures H, in billions of dollars, can be modeled b
dmitriy555 [2]

Answer:

2019.

Step-by-step explanation:

We have been given that for the years from 2002 and projected to 2024, the national health care expenditures H, in billions of dollars, can be modeled by H = 1,500e^{0.053t} where t is the number of years past 2000.

To find the year in which national health care expenditures expected to reach $4.0 trillion (that is, $4,000 billion), we will substitute H=4,000 in our given formula and solve for t as:

4,000= 1,500e^{0.053t}

\frac{4,000}{1,500}=\frac{ 1,500e^{0.053t}}{1,500}

\frac{8}{3}=e^{0.053t}

e^{0.053t}=\frac{8}{3}

Take natural log of both sides:

\text{ln}(e^{0.053t})=\text{ln}(\frac{8}{3})

0.053t\cdot \text{ln}(e)=\text{ln}(\frac{8}{3})

0.053t\cdot (1)=0.9808292530117262

\frac{0.053t}{0.053}=\frac{0.9808292530117262}{0.053}

t=18.506212320

So in the 18.5 years after 2000 the expenditure will reach 4 trillion.

2000+18.5=2018.5

Therefore, in year 2019 national health care expenditures are expected to reach $4.0 trillion.

7 0
3 years ago
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