X^2+4+10=0
x=(-4(+/-)root16-40)/2, so we now know that the zeros are imaginary, because you can't square root a negative number and 16-40 is -24
so the two roots are…
-2+iroot6 and -2-iroot6

7 0

3 years ago

Read 2 more answers

Plz help will give brainliest!

natita [175]

Answer:

8.7

Step-by-step explanation:

4 0

2 years ago

Find the maximum value or minimum value for the function f(x) = 0.15(x + 1)² - 3.

il63 [147K]

Answer:

The minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

Step-by-step explanation:

Given function is $f(x) = 0.15(x + 1)^2 - 3$

We need to find the maximum value or the minimum value for the function.

Now, differentiate $f(x) = 0.15(x + 1)^2 - 3$ w.r.t $x$ .

$f'(x) =\frac{d}{dx}(0.15(x + 1)^2 - 3)\\f'(x)=\frac{d}{dx}(0.15(x+1)^2-\frac{d}{dx}(3)\\$

$f'(x)=2\times 0.15(x+1)\frac{d}{dx}(x+1)-0\\f'(x)=0.3(x+1)(1)\\f'(x)=0.3(x+1)$

Now, we will equate $f'(x)=0$ to find critical point.

$0.3(x+1)=0\\x=-1$

Plug this critical point in to the function $f(x) = 0.15(x + 1)^2 - 3$ we get,

$f(-1) = 0.15(-1 + 1)^2 - 3\\f(-1)=-3$

Also, $f''(x)=0.3$ which is positive, We have minimum value.

So, the minimum value for $f(x) = 0.15(x + 1)^2 - 3$ is $-3$ .

7 0

3 years ago