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3 years ago

Slope of (0,1) and (5,-3)

Mathematics

1 answer:

victus00 [196]3 years ago

5 0

Answer:

Your answer is $\frac{-4}{5}$

Step-by-step explanation:

1. Slope formula is $\frac{y2-y1}{x2-x1}$

2. Substitute: $\frac{-3-1}{5-0}$

3. Subtract: $\frac{-4}{5}$

Hope this helps. Leave a comment if you have questions!

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The system of equations that represent the situation are as follows;

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What is the factored term form of 2x^3+4x^2-x

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The correct works are:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$ .
$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

<h3>Function Notation</h3>

The function is given as:

$Blue(s) = 2s^2 + 3$

The interpretation when Steven is asked to calculate Blue(s + h) is that:

Steven is asked to find the output of the function Blue, when the input is s + h

So, we have:

$Blue(s + h) = 2(s + h)^2 + 3$

Evaluate the exponent

$Blue(s + h) = 2(s^2 + 2sh + h^2) + 3$

Expand the bracket

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

So, the correct work is:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

<h3>Simplifying Difference Quotient</h3>

In (a), we have:

$Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3$

$Blue(s) = 2s^2 + 3$

The difference quotient is represented as:

$\frac{f(x + h) - f(x)}{h}$

So, we have:

$\frac{Blue(s + h) - Blue(s)}{h} = \frac{2s^2 + 4sh + 2h^2 + 3 - 2s^2 - 3}{h}$

Evaluate the like terms

$\frac{Blue(s + h) - Blue(s)}{h} = \frac{4sh + 2h^2}{h}$

Evaluate the quotient

$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

Hence, the correct work is:

$\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h$

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