Consider the equation - 3x2 + 27x - 63 = 0. What is the discriminant of the equation?

Mathematics

1 answer:

poizon [28]2 years ago

5 0

Answer:

answer is 19/9.

Step-by-step explanation:

give me brielient.

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If y varies directly as x, and y is 12 when x is 1.2, what is the constant of variation for this relation? A) 1 B) 10 C) 10.8 D)

Anit [1.1K]

$\bf \begin{array}{cccccclllll}&\\
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\&\\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\&\\
y&=&{{ k}}&\cdot&x&\\
&\\
&& y={{ k }}x&\\
\end{array}& \\\\\\
\textit{we know that }
\begin{cases}
y=12\\
x=1.2
\end{cases}\implies y=kx\implies 12=k\cdot 1.2$

solve for "k" to find "k", or the "constant of variation"

6 0

3 years ago

Read 2 more answers

Va rog e urgent am nevoie de raspuns

Lady_Fox [76]

Answer: 4

<u>Explanation:</u>

f(x) = 2x - 1

f(√2) = 2√2 - 1

f(1) = 2(1) - 1

= 2 - 1

= 1

f(√3) = 2√3 - 1

*******************************************************

$\frac{f(\sqrt{2})-f(1)}{\sqrt{2}-1} +\frac{f(\sqrt{3})-f(\sqrt{2})}{\sqrt{3}-\sqrt{2}}$

= $\frac{(2\sqrt{2}-1)-1}{\sqrt{2}-1} +\frac{(2\sqrt{3}-1)-(2\sqrt{2}-1)}{\sqrt{3}-\sqrt{2}}$

= $\frac{2\sqrt{2}-2}{\sqrt{2}-1} +\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

= $\frac{2\sqrt{2}-2}{\sqrt{2}-1}(\frac{\sqrt{2}+1}{\sqrt{2}+1})+\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}})$