A cylinder has a radius of 12.2 inches and a height of 6.1 inches.

Let f be the function defined as follows.

Debora [2.8K]

Y = f(x) = 8x^2 - 2x + 3
a.) dy/dx = 16x - 2

b.) dy = (16x - 2)dx
dy = (16(2) - 2)(2 - 1.97) = 0.03(32 - 2) = 0.03 * 30 = 0.9
dy = 0.9

c.) (8(2)^2 - 2(2) + 3) - (8(1.97)^2 - 2(1.97) + 3) = 31 - 30.1072 = 0.8928
Δy = 0.8928

4 0

4 years ago

PLS HELP! Will give 25 points!

RSB [31]

Answer and explanation:

Given: Belinda wants to invest $1,000. The table below shows the value of her investment under two different options for three different years:

Number of years 1 2 3

Option 1 (amount in dollars) 1100 1200 1300

Option 2 (amount in dollars) 1100 1210 1331

To find:

Part A: What type of function, linear or exponential, can be used to describe the value of the investment after a fixed number of years using option 1 and option 2?

Part B: Write one function for each option to describe the value of the investment f(n), in dollars, after n years.

Part C: Will there be any significant difference in the value of Belinda's investment after 20 years if she uses option 2 over option 1?

Solution:

Part A: Linear and exponential functions can be used to describe the value of the investment after a fixed number of years using option 1 and option 2, respectively.

Part B: (n=n+100) and (n=n+100x) are the functions for each option to describe the value of the investment f(n), in dollars, after n years.

Part C: Yes, there will be a significant difference of 1900 in the value of Belinda's investment after 20 years if she uses option 2 over option 1.

Part A:

In the case of option 1, the linear function can be used to describe the value of the investment after a fixed number of years. This is because, in option one, the amount increases by a fixed amount every year.

In the case of option 2, the exponential function can be used to describe the value of the investment after a fixed number of years. This is because, in option 2, the amount increase is higher than last year.

Part B:

For option 1, the function is

For option 2, the function is

Here, x is the increase in amount every consecutive year.

Part C:

After 20 years, the amount from option 1 would be 3000 and the amount from option 2 would be 4900. Thus, there is a difference between 1900.

Therefore,

Part A: Linear and exponential functions can be used to describe the value of the investment after a fixed number of years using option 1 and option 2, respectively.

Part B: (n=n+100) and (n=n+100x) are the functions for each option to describe the value of the investment f(n), in dollars, after n years.

Part C: Yes, there will be a significant difference of 1900 in the value of Belinda's investment after 20 years if she uses option 2 over option 1.

Hope this helps

8 0

2 years ago

Solve by substitution x-y=-11 y+7=-2x

Bess [88]

x - y = - 11 .............( 1 )

y + 7 = - 2x ...............( 2 )

from equation ( 1 )

x - y = - 11

x = -11 + y ...........( 3)

putting x in equation ( 2 )

y + 7 = - 2 x

y + 7 = -2 ( -11 + y ) y + 7 = 22 - 2 y y + 2 y =22 - 7

3 y = 15 y = 15 / 3

putting value of y in equation 3

x = -11+ ( 13 /5 ) x = -33 /3 + 15 / 3 ( l.c.m)

x = -17 / 5

check

x = -11 + y

- 17 / 5 = -11 + 13 / 5

-17 /5 = -17 / 5

4 0

3 years ago

Find the value of X in each of the following2x-7=13 3x+4=25

Law Incorporation [45]

Answer:

2x-7=13 X=10

3x+4=25 X=7

2/3x=8 X=12

3/2x=6 X=4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The hawaiian alphabet has 12 letters. How many permutations are possible for five of these letters

guajiro [1.7K]

Answer: The answer is 50980

Step by Step Explanation:

First, temporarily assume that two letters with I are different, call them i 1 and i 2. Three "a" are also called as a 1, a 2 and a 3, and two h as h 1 and h 2. Then there are 11 * 10 * 9 * 8 * 7 = 55440 possible "words" (one of 11 is the first letter, 10 is the second, and so on). But because equal letters do the same "words," some "words" were counted twice or more. We have to deduct the number of "parasitic" counts although it is fairly small. The words that counted more than once are divided into many disjoint sets: 1) with two I but without repetitions of a and h; 2) with two h but without repetitions of a and I 3) with two a but without repetitions of I and h; 4) with three a but without repetitions of I and h; 5) with two I and two a's; 6) two i's and tree a's; 7) two i's and two h's 8) two h's and two a's; 9) two h's and one tree. The first category includes terms counted twice and its scale is (5 * 4) * (6 * 5 * 4) = 2400 (the first I stays at one of the 5 positions, the second at one of the 4, then 11-2i-1h-2a=6). So we have 2400/2 = 1200 to subtract. Group 2 gives -600 as well, and group 3 also. Group 4 gives * (6 * 5) = 1800 (5 * 4 * 3), and the terms are counted 6 times, -300. Groups 5, 7 , 8: 5 * 4 * 3 * 2 * 6 = 720 and counted four times, therefore -180. Group 6 and 9: 5 * 4 * 3 * 2 * 1 = 120, with 12 counts, -10. Altogether -(1200 * 3 + 300 + 180 * 3 + 10 * 2) = -4460.The answer will be 55440-4460 = 50980.

3 0

4 years ago