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2 years ago

Can somoene help me!! this is due today!

Mathematics

1 answer:

vodka [1.7K]2 years ago

7 0

Answer:

The answer would be f(x)={4x/9}

Step-by-step explanation:

f(x)=4({1}{3})^{2x}

Apply exponent rule:

f(x)=4({1^2}{3^2})^x

f(x)=4({1}{9})^x

the required function is f(x)=4({1}{9})^x

If the function is f(x)=4(\frac{1}{3})^2x

Apply exponent rule:

f(x)=4({1^2}{3^2})x

f(x)=4({1}{9})x

f(x)={4x}{9}

Hence, the required function is f(x)={4x}{9}

I hope this helped. I am sorry if you get this wrong.

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A physical education teacher plans to divide the seventh graders at Wilson Middle School into teams of equal size of year-ending

Crank

Step-by-step explanation:

Consider the provided information.

The total number of students are: 24+26+21=71

He wants each team will have the same number of students between 5 and 9.

The number between 5 to 9 are 6, 7 and 8.

The number 71 is a prime number so we need to divide the 71 students in group so that minimum students left without team.

Dividing 6 students in each group.

Number of teams: $\frac{71}{6}$ = 11 teams and 5 students.

Dividing 7 students in each group.

Number of teams: $\frac{71}{7}$ = 10 teams and 1 students.

Dividing 8 students in each group.

Number of teams: $\frac{71}{8}$ = 8 teams and 7 students.

Thus, we can not divide 71 students into equal teams, but 70 divided by 7 gives the least students left out i.e 10 teams each with 7 students and only 1 student will remain at last.

6 0

3 years ago

I don’t know how to do this

stich3 [128]

To check for continuity at the edges of each piece, you need to consider the limit as $x$ approaches the edges. For example,

$g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}$

has two pieces, $2x+5$ and $x^2-10$ , both of which are continuous by themselves on the provided intervals. In order for $g$ to be continuous everywhere, we need to have

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)$

By definition of $g$ , we have $g(-3)=2(-3)+5=-1$ , and the limits are

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1$

$\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1$

The limits match, so $g$ is continuous.

For the others: Each of the individual pieces of $f,h$ are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.