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garik1379 [7]
2 years ago
11

Help please i need help on this question

Mathematics
1 answer:
cupoosta [38]2 years ago
4 0

9514 1404 393

Answer:

  A: range

  B: 0 ≤ f(x) ≤ 6

Step-by-step explanation:

<h3>Part A</h3>

As is often the case in multiple-part multiple-choice questions, the answer to the first part is found in the second part. The "<em>range</em>" is the set of all possible output values.

__

<h3>Part B</h3>

The range is the vertical extent of the graph. This graph goes from a minimum of 0 to a maximum of 6 and includes every value in between. The range is ...

  0 ≤ f(x) ≤ 6

_____

<em>Additional comment</em>

You should discuss this question with your teacher. The left inequality symbol on the bottom line is pointed the wrong way.

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The answer is 36/18.

Hope this helps.

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Which of the following phrases are inequalities?
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What is m in equation 2m=12
Likurg_2 [28]
M is equal to 6 because 6 times 2 would equal 12
7 0
3 years ago
. Consumer Awareness In 1986, the average cost of a new midsize four-door sedan was $9000. In 1991, the average cost was $12,000
eduard

<u>Answer-</u>

\boxed{\boxed{y=9000(1.0592)^x}}

<em>And cost of a car in the year 2000 will be </em><em>$20134.18</em>

<u>Solution-</u>

Let's assume,

x = number of years after 1986

y = average cost of sedan in dollar

The exponential model that will model the scenario will be in the form of,

y=ab^x

where a and b are constants

As given that, in 1986 average cost of sedan was $9000 and in 1991 average cost of sedan was $12000

So, the points (0, 9000) and (5, 12000) will satisfy or lie on the exponential curve.

Putting (0, 9000) in the equation,

\Rightarrow 9000=ab^0

\Rightarrow a\times 1=9000

\Rightarrow a=9000

Now, the equation becomes y=9000b^x

Putting (5, 12000) in this equation,

\Rightarrow 12000=9000b^5

\Rightarrow b^5=\dfrac{12000}{9000}

\Rightarrow b=\sqrt[5]{\dfrac{4}{3}}

\Rightarrow b=1.0592

Putting the values,

y=9000(1.0592)^x

As we have to calculate the cost of sedan in 2000, so putting x=14(as 2000-1986=14),

y=9000(1.0592)^{14}=20134.18

Therefore, cost of a car in the year 2000 will be $20134.18

8 0
3 years ago
Calculate the value of the following factorial expressions.
never [62]

Answer:

(a) 7

(b) 5040

(c) 336

(d) 132                          

Step-by-step explanation:

We have to calculate the value of given expression

(a) \frac{7!}{6!}

\frac{7!}{6!}=\frac{7\times 6!}{6!}=7

(b) \frac{10!}{6!}=\frac{10\times 9\times 8\times 7\times 6!}{6!}=10\times 9\times 8\times 7=5040

(c) \frac{8!}{5!}=\frac{ 8\times 7\times 6\times 5!}{5!}= 8\times 7\times 6=336

(d) \frac{12!}{10!}=\frac{ 12\times 11\times 10!}{10!!}= 12\times 11=132

6 0
3 years ago
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