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2 years ago

Given the graph below, which of the following represents the domain of the function?

Mathematics

1 answer:

krek1111 [17]2 years ago

7 0

Do you have the graph?

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Help with this question

Marat540 [252]

Lxwxh using your numbers its 3x3x3 or 3³ so your answer is 27

4 0

3 years ago

Read 2 more answers

Question 2

frosja888 [35]

Answer:

(50,55,60,....)

Step-by-step explanation:

Given that a health insurance plan has a $50 enrollment fee and a copayment of $5 for each doctor visit.

The function f(x) = 5x + 50 gives the cost of x doctor visits

Here x is the no of visits.

No of visits is an independent variable while f(x) total cost is a dependent variable on x

slope = 5 which indicates for each visit extra, extra cost is 5 dollars

Enrollment fee = 50 = fixed cost

Hence domain would be

x =0,1,2,3.....

So range is

(50,55,60,....)

3 0

3 years ago

2. Si 3x + 2 = y, entonces (3x + 2) + 5 = y + 5*

jeyben [28]

Answer:

26.9y2

Step-by-step explanation:

8 0

3 years ago

I don’t understand this help please

Elodia [21]

Answer:

Step-by-step explanation:

The first simplification we can make is to replace 6^0 with 1.

The first rules of exponents we can apply are ...

(a·b)^c = a^c·b^c . . . . . . similar to the distributive property

 (a^b)^c = a^(b·c)

This reduces your expression to ...

$(2^8\cdot 3^{-5})^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot 2^{28}=2^{8(-2)}\cdot 3^{-5(-2)}\cdot\dfrac{3^{-2(4)}}{2^{3(4)}}\cdot 2^{28}\\\\=2^{-16}\cdot 3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot 2^{28}$

Now we can apply another two rules of exponents:

 (a^b)(a^c) = a^(b+c)

 1/a^b = a^-b

Using these, we have ...

$=2^{-16-12+28}\cdot 3^{10-8}\\\\=2^0\cdot 3^2\\\\=1\cdot 9=9$

The value of the expression is 9.

3 0

3 years ago

Caroline invested $1100 in an account that pays 2% interest

Nadya [2.5K]

Answer: 1480.5

Step-by-step explanation:

8 0

2 years ago