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3 years ago

Please help me now :c

Mathematics

1 answer:

nadya68 [22]3 years ago

8 0

On Monday he reads 2 pages, on Tuesday he reads 6 pages (triple of 2: 2 + 2 + 2 = 6), on Wednesday he reads 18 pages (triple of 6: 6 + 6+ 6 = 18), on Thursday he reads 54 pages (triple of 18: 18 + 18 + 18 = 54).

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For what value of constant c is the function k(x) continuous at x = 0 if k =

nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x) is continuous, we take the limit of the parameter as follows:

$\mathbf{ \lim_{x \to 0^-} k(x) = \lim_{x \to 0^+} k(x) = 0 }$

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= c }$

Provided that:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }$

Using l'Hospital's rule:

$\mathbf{\implies \lim_{x \to 0} \ \ \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}= \lim_{x \to 0} sec \ x \ tan \ x = 0}$

Therefore:

$\mathbf{\implies \lim_{x \to 0 } \ \ \dfrac{sec \ x - 1}{x}=0 }$

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0

2 years ago

Believe this is worth 7 pts please help!

salantis [7]

Answer:

-12

Step-by-step explanation:

The average rate of change is found by

f(4) - f(2)

---------------

4-2

32 - 56

----------

4-2

-24

------

-12

7 0

3 years ago

What is the quotient? StartFraction n + 3 Over 2 n minus 6 EndFraction divided by StartFraction n + 3 Over 3 n minus 9 EndFracti

shepuryov [24]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Given the expression:

$\frac{n+3}{2n-6} \div \frac{n+3}{3n-9}$

We are to find the quotient as shown:

$= \frac{n+3}{2n-6} \div \frac{n+3}{3n-9} \\\\= \frac{n+3}{2n-6} \times \frac{3n-9}{n+3} \\\\= \frac{n+3}{n+3} \times \frac{3n-9}{2n-6}\\\\= 1 \times \frac{3(n-3)}{2(n-3)}\\\\= 1 \times \frac{3}{2} \\\\= \frac{3}{2}$