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Levart [38]
4 years ago
6

Difference quotient

Mathematics
1 answer:
Nina [5.8K]4 years ago
6 0
f(x)=5x^2-5x-6
\\ \quad \\

\begin{cases}
f(\boxed{x+h})=5(\boxed{x+h})^2-5(\boxed{x+h})-6
\end{cases}\qquad thus
\\ \quad \\
\cfrac{f(x+h)-f(x)}{h}\qquad \textit{will be then}
\\ \quad \\
\cfrac{[5({x+h})^2-5({x+h})-6]\quad -\quad [5x^2-5x-6]}{h}
\\ \quad \\
\cfrac{[5(x^2+2xh+h^2)-5(x+h)-6]-[5x^2-5x-6]}{h}


\cfrac{\underline{5x^2}+10xh+5h^2\underline{-5x}-5h\underline{-6}\underline{-5x^2}\underline{+5x}\underline{+6}}{h}\impliedby \textit{canceling those ones}
\\ \quad \\
\cfrac{10xh+5h^2-5h}{h}\impliedby \textit{common factor}
\\ \quad \\
\cfrac{5\underline{h}(2x+h-1)}{\underline{h}}

and surely, you'd know what that is
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  10.7 years

Step-by-step explanation:

The decay equation can be written as ...

  remaining = initial × (1/2)^(t/(half-life))

Filling in the given values, we can solve for t.

  0.100 = 1.35 × (1/2)^(t/2.86)

  0.100/1.35 = (1/2)^(t/2.86) divide by 1.35

Taking logs transforms this to a linear equation:

  log(0.100/1.35) = (t/2.86)log(1/2)

Since log(a/b) = -log(b/a), we can multiply both sides by -1 and simplify the logs a bit.

  log(1.35/.1) = t·(log(2)/2.86)

  2.86·log(13.5)/log(2) = t ≈ 10.7 . . . . years

The decay time is about 10.7 years.

6 0
4 years ago
For which of the following is x = 5 NOT a solution
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A) x= -9

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Step-by-step explanation:

6 0
3 years ago
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Every year,the value of an antique vase appreciates by 20% of it's value in the previous year if the value of case was 180000 in
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Answer: Every year, the value of an antique vase appreciates by 20% of its value in the previous year. If the value of the vase was $180 000 in 2012, its value in 2010 was $125000

Step-by-step explanation:

Given the appreciation rate r, the initial value I, the final value V, and the period of appreciation n, the formula for appreciation is given by:

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In our case, we need to work out the value of n.

We do this as follows:

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Example of attributes are religion, educational level, gender, color of hair,etc. A change in attribute can only be expressed qualitatively.

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