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NISA [10]
3 years ago
5

Can derivatives be used to measure the the slope of the tangent line at x=a

Mathematics
1 answer:
Pie3 years ago
6 0
Yes. In fact, that is basically the definition of a derivative. It is the instantaneous rate of change of a function. 

For example, picture the graph of the following function:

f(x) = x^2

The slope is constantly changing at every x-value, so to find  the slope at x=a, we find the  derivative of the  function.

f'(x)=2x

Once we have the derivative, simply plug in a for x to find the slope of the line tangent to f(x) at x=a.

For example, at x=5:

f'(5)=2(5)

The slope of f(x) at x=5 is 10.
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The slope in the linear equation y=12x+5 is _[blank]_.
Kazeer [188]
The slope is 12x, remember the slope is the number value before the variable!
7 0
2 years ago
Will mark brainlits!!!!!!!!!!!!!!!!!!! help me plz owo ASAP
Fudgin [204]

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the area of a square with a side length of 9x  

Step-by-step explanation:

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3 years ago
Which equation is equivalent to the given equation?
madam [21]

Answer:

x² - x - 12 = 0

Step-by-step explanation:

0 = -x² + x + 12

→ Add x² to both sides

x² = x + 12

→ Minus x from both sides

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4 0
2 years ago
In a sample of 679 new websites registered on the Internet, 42 were anonymous (i.e., they shielded their name and contact inform
Semmy [17]

Answer:

95% Confidence interval:  (0.0429,0.0791)      

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 679

Number of anonymous websites, x = 42

\hat{p} = \dfrac{x}{n} = \dfrac{42}{679} = 0.0618

95% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.05} = \pm 1.96

Putting the values, we get:

0.0618\pm 1.96(\sqrt{\dfrac{0.0618(1-0.0618)}{679}}) = 0.0618\pm 0.0181\\\\=(0.0429,0.0791)

is the required confidence interval for proportion of all new websites that were anonymous.

8 0
3 years ago
Round 20045 to 4 significant figures.
tresset_1 [31]

Answer:

2005

Step-by-step explanation:

4 0
3 years ago
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