PLEASE HELP ME ANSWER THIS

1 answer:

8 0

The function is neither. I just put the function into a calculator. Plus a linear function doesn't have exponents. A quadratic function is ax^2+bx+c so it can't be quadratic.

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Good morning....I guess??

Simora [160]

Answer:

goodnight...efhfcbrbxbcjbxbxbfj

5 0

4 years ago

Read 2 more answers

A sequence is defined recursively using the equation f(n + 1) = f(n) – 8. If f(1) = 100, what is f(6)?

BARSIC [14]

If the sequence f is defined as

$\mathsf{f(n+1)=f(n)-8}$

then

$\mathsf{f(n+1)-f(n)=-8}$

As you can see above, the difference between two consecutive terms gives us a constant.

So f is actually an arithmetic progression (AP), and it has (– 8) as its common difference:

$\mathsf{f(n)=f(1)+(n-1)\cdot d}\\\\ \textsf{where }\mathsf{d=-8.}$

• If f(1) = 100, then

$\mathsf{f(6)=f(1)+(6-1)\cdot (-8)}\\\\ \mathsf{f(6)=100+5\cdot (-8)}\\\\ \mathsf{f(6)=100-40}\\\\ \boxed{\begin{array}{l} \mathsf{f(6)=60} \end{array}}\qquad\checkmark$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2190444

I hope this helps. :-)

3 0

3 years ago

Perform the indicated operation and express the result as a simplified complex number. I need help with 35

skelet666 [1.2K]

Given:-

$\frac{3+4i}{2-i}$

To find:-

The simplified form.

At first we take conjucate and multiply below and above.

The conjucate is,

$2+i$

So now we multiply. we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}$

Now we simplify. so we get,

$\frac{3+4i}{2-i}\times\frac{2+i}{2+i}=\frac{6+3i+8i+4(i)^2}{2^2-i^2}$

We know the value of,

$i^2=-1$

Substituting the value -1. we get,

$\begin{gathered} \frac{6+3i+8i+4(i)^2}{2^2-i^2}=\frac{6+3i+8i+4(-1)_{}^{}}{2^2-(-1)^{}} \\ \text{ =}\frac{6-4+11i}{2+1} \\ \text{ =}\frac{2+11i}{3} \end{gathered}$

So now we split the term to bring it into the form a+ib. so we get,