If -nx2 + tx + c = 0, what is x equal to?

1 answer:

5 0

$f(x)=-nx^2+tx+c=0\\\\
\Delta=t^2-4\cdot(-n)\cdot c=t^2+4cn\\\\
1.\\
 \Delta0\\
\sqrt{\Delta}=\sqrt{t^2+4cn}\\
x_1=\frac{-t-\sqrt{t^2+4cn}}{2\cdot(-n)}=\frac{t+\sqrt{t^2+4cn}}{2n}\\
x_2=\frac{-t+\sqrt{t^2+4cn}}{2\cdot(-n)}=\frac{t-\sqrt{t^2+4cn}}{2n}\\$

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enyata [817]

Answer:
f(8) = 192

Explanation:
f(x) varies directly with x²
This means that:
f(x) = kx²
where k is the constant of proportionality.

Now, we are given that:
f(x) = 75 at x = 5
Substitute to get k as follows:
f(x) = kx²
75 = k(5)²
75 = 25k
k = 3

This means that the equation is:
f(x) = 3x²

Now, to find f(x) at x = 8, all we have to do is substitute with x = 8 in the equation as follows:
f(8) = 3*(8)²
f(8) = 3*64
f(8) = 192

Hope this helps :)

8 0

4 years ago

Segment bc is parallel to De. Use the proportion from a B to BD to find the missing length of CE

Alla [95]

Answer:

CE = 6

Step-by-step explanation:

we use the proportion relationship:

$\frac{AB}{BD} = \frac{AC}{CE\\}\\then$