3 years ago

AP calculus HW Need help on #61

1 answer:

7 0

Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/∆x

Crossing the 1s , the 2xs, and the x^2s out, we get

(2x∆x+∆x^2-2∆x)/∆x

Dividing the ∆x, we get
2x+∆x-2.

Making the limit of ∆x=0, we get 2x-2.

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kaheart [24]

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Stells [14]

Answer: The slope of the line is $\frac{1}{2}$

Step-by-step explanation:

The slope of a line can be calculated with the following formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

Then, given the points $(-3, 5)$ and $(1, 7)$ , you can identify that:

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Knowing these values you can substitute them into the formula for calculate the slope:

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Finally, evaluating, you get that the slope of that line is:

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8 0

3 years ago

When I solved this equation I got x=0.5

qwelly [4]

You might have made an error the first time you solved for x. I got x = -0.5.

When you have your log base 4, the way you cancel that out is by making 4 the base on both sides, so you get 4^(log4) to reduce to 1, and you're left with:

2x + 3 = 4^(1/2) ... Simplify

2x + 3 = 2

2x = -1

x = -1/2

If you plug that back in, everything checks out. Maybe double check your use of logarithm/exponent properties?

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