4. List 3 rational numbers between 3 and 3.9.

(a) Use the Quotient Rule to differentiate the function f(x)=tan(x)-1/sec(x). f'(x)=

yKpoI14uk [10]

Answer:

(a) sin(x) + cos(x)

(b) sin(x) + cos(x)

(c) Both answers are equivalent

Step-by-step explanation:

(a) The given function is:

$f(x) = \frac{tan(x)-1}{sec(x)}$

According to the quotient rule:

$f(x) = \frac{g(x)}{h(x)}\\f'(x)= \frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$

Applying the quotient rule:

$f(x) = \frac{tan(x)-1}{sec(x)}\\f'(x)=\frac{sec^2(x)*sec(x)-(tan(x)-1)*sec(x)tan(x)}{sec(x)^2}\\f'(x)=\frac{sec^3(x)-sec(x)tan^2(x)+sec(x)tan(x)}{sec(x)^2}\\f'(x)=sec(x)+\frac{tan(x)-tan^2(x)}{sec(x)}\\ \frac{tan(x)}{sec(x)}=sin(x) \\f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\$

This can be simplified to:

$f'(x)=sec(x)+sin(x)-sin(x)tan(x)\\f'(x) = \frac{1}{cos(x)}+sin(x)-\frac{sin^2(x)}{cos(x)}\\f'(x)=\frac{1+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{sin^2(x)+cos^2(x)+sin(x)cos(x)-sin^2(x)}{cos(x)}\\f'(x)=\frac{cos^2(x)+sin(x)cos(x)}{cos(x)}\\ f'(x)=sin(x) +cos(x)$

(b) Simplifying in terms of sin(x) and cos(x):

$f(x) = \frac{tan(x)-1}{sec(x)}\\f(x)=\frac{\frac{sin(x)}{cos(x)}-1 }{\frac{1}{cos(x)} } \\f(x)=sin(x)-cos(x)\\f'(x) = cos(x)+sin(x)$

(c) As proven above, both answers are equivalent.

4 0

3 years ago

G.SRT.1 In the coordinate plane, quadrilateral ABCD has vertices A (1, 1), * 1 point B (5,1), C (5,9), and D (1, 9). Quadrilater

Natalka [10]

Based on the vertices of quadrilateral ABCD and Quadrilateral FGHJ, A. The scale factor is 1/2, and the center of dilation is point A.

<h3>What is the scale factor?</h3>

First, note the vertices of Quadrilateral FGHJ:

F(1,1) G(3 ,1) J (1 , 5) H(3 , 5)

Find the lengths of the sides of both quadrilaterals:

FG = (3, 1) - (1, 1) AB = (5,1) - (1, 1)

= (2,0) = (4, 0)

FH = (3, 5) - (1, 1) AC = (5,9) - (1, 1)

= (2,4) = (4, 8)

AB is twice the size of FG.

AC is twice the size of FH

This shows that quadrilateral FGHJ is 1/2 the size of quadrilateral ABCD.

Find out more on dilation at brainly.com/question/10253650.

#SPJ1

4 0

2 years ago

Assume the number of pounds of chocolate consumed in a year is equally likely between 20 and 100 pounds. what is the probability

kramer

Because the probability is uniform (equally likely) between 20 and 100 pounds, the probability distribution is uniform between 80 and 100 pounds, with the value 1/(100-20) = 1/80 chocolates/pound.

Let x = random variable that represents pounds of chocolate consumed.
Therefore
P(x < 60) = (60 - 20)*(1/80) = 1/2.

Answer: The probability is 1/2.

4 0

3 years ago

Amy earns $9.25/h. She worked 8 hours on Monday, 10 hours on Tuesday, 12 hours on Wednesday, 6 hours on Thursday, 6 hours on Fri

Nana76 [90]

A. She worked 6 hrs overtime
B. $9.25 x 40 = 370
6 x $13.88 = 83.28

Her weeks pay is $453.28

I hope that answered you question and that it’s right !

8 0

3 years ago

Given the function f(x) = 5x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

Debora [2.8K]

$f(x)=5x\\\frac{f(1)-f(0)}{1-0}\\\frac{5-0}{1}\\5\\\\\frac{f(3)-f(2)}{3-2}\\\frac{15-10}{1}\\5$

A: 5; 5

B: 1; they are the same

3 0

3 years ago