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2 years ago

Fill in the blank to complete the equation below.

Mathematics

2 answers:

stiv31 [10]2 years ago

8 0

The answer is 6
i just did the work

Vika [28.1K]2 years ago

6 0

I think the answer is 6

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If the function f(x) has a domain of (g,h) and a range of (j,k), then what is the domain and range of g(x) = m(f(x)] + p?

sertanlavr [38]

Answer:

$Dom\{g(x)\} = Dom\{f(x)\} = (g, h)$ .

$Ran \{g(x)\} = (m\cdot j+p,m\cdot k +p)$

Step-by-step explanation:

From Mathematics we remember that the domain of a functions corresponds to the set of values of the independent variable ( $x$ in this case) so that images exist and the range of a function is the set of images.

In this case, we know the domain and range of $f(x)$ and we must find the domain and range of $g(x)$ .

Domain

The domain of $g(x)$ is the domain of $f(x)$ . That is, $Dom\{g(x)\} = Dom\{f(x)\} = (g, h)$ .

Range

We have to define the bounds of the range of $g(x)$ , given that range $f(x)$ is modified by streching and horizontal translation operations:

Lower bound ( $f(x) = j$ )

$g(x) = m\cdot j +p$

Upper bound ( $f(x) = k$ )

$g(x) = m\cdot k +p$

In consequence, the range of $g(x)$ is $Ran \{g(x)\} = (m\cdot j+p,m\cdot k +p)$

3 0

3 years ago

Problem 4. Determine which of the following properties: 1. Memoryless 2. time invariant 3. linear 4. causal hold, and which do n

LenaWriter [7]

-24 G g Gg f f f f f g g g g f f.

7 0

3 years ago

HI. I really Need help. I will Mark Brainliest!

Reptile [31]

Answer:

the first 1 and the last 1 i think

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

In 2018, the population will grow over 700,000.The population of a city in 2010 was 450,000 and was growing at a rate of 5% per

Firlakuza [10]

Answer:

The year is 2020.

Step-by-step explanation:

Let the number of years passed since 2010 to reach population more than 7000000 be 'x'.

Given:

Initial population is, $P_0=450,000$

Growth rate is, $r=5\%=0.05$

Final population is, $P=700,000$

A population growth is an exponential growth and is modeled by the following function:

$P=P_0(1+r)^x$

Taking log on both sides, we get:

$\log(P)=\log(P_0(1+r)^x)\\\log P=\log P_0+x\log (1+r)\\x\log (1+r)=\log P-\log P_0\\x\log(1+r)=\log(\frac{P}{P_0})\\x=\frac{\log(\frac{P}{P_0})}{\log(1+r)}$

Plug in all the given values and solve for 'x'.

$x=\frac{\log(\frac{700,000}{450,000})}{\log(1+0.05)}\\x=\frac{0.192}{0.021}=9.13\approx 10$

So, for $x > 9.13$ , the population is over 700,000. Therefore, from the tenth year after 2010, the population will be over 700,000.

Therefore, the tenth year after 2010 is 2020.

8 0

2 years ago

Help please (will give brainliest)

lorasvet [3.4K]

Step 1, all the exponents are multiplied by 2

3 0

2 years ago