Heyyy ummm... help??

Mathematics

2 answers:

attashe74 [19]3 years ago

8 0

Answer:

omgg yess

Step-by-step explanation:

atroni [7]3 years ago

7 0

The statements true for the given function are:

$f(0) = \frac{3}{2}$
$f(4) = \frac{7}{2}$

I tried them all and these two were correct, their solutions are as follows:

= f(x) = 1/2x + 3/2

= f(0) = 1/2 × 0 + 3/2

= f(0) = 0 + 3/2

= f(0) = 3/2

= f(x) = 1/2x + 3/2

= f(4) = 1/2 × 4 + 3/2

= f(4) = 2 + 3/2

= f(4) = 4+3/2

= f(4) = 7/2

So, that's how these two are correct.

You might be interested in

If c = 4 and d = 5, find c:d.

Julli [10]

Would it not be 4:5 as a ratio?

8 0

3 years ago

Read 2 more answers

Limit question: lim x-->pi ((e^sinx)-1)/(x-pi)

aleksandr82 [10.1K]

$\displaystyle\lim_{x\to\pi}\dfrac{e^{\sin x}-1}{x-\pi}$

Notice that if $f(x)=e^{\sin x}$ , then $f(\pi)=e^{\sin\pi}=e^0=1$ . Recall the definition of the derivative of a function $f(x)$ at a point $x=c$ :

$f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

So the value of this limit is exactly the value of the derivative of $f(x)=e^{\sin x}$ at $x=\pi$ .

You have

$f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1$