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Umnica [9.8K]
2 years ago
10

Find the quotient of the following: f(x)=2x-5, g(x) 1-x

Mathematics
1 answer:
AfilCa [17]2 years ago
6 0
What are the answer choices
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Help me with this plzzzzzzzz the number under one is three
Black_prince [1.1K]

Answer:

sorry.......dud.....

Step-by-step explanation:

hope some one will help you........

4 0
3 years ago
For Restaurant A, the owner wants to advertise a good cleanliness score to his customers. Which measure would make his score see
Semenov [28]

Answer:

median

Step-by-step explanation:

the median would make the score as good as possible, because there could be possible outliers. this really depends on if there are outliers or not. here is an example of what i need:

say you are trying to show your parents your grades:

here are your scores:

38, 88, 90, 91, 93, 95, 100

the median would be 91, which looks better.

the mean would be 85.

even though a <em>majority</em> of your grades are good, the one 38 makes the average go down. therefore, the median is a better advertiser.

8 0
4 years ago
Plz help I need to know please.....
jarptica [38.1K]

101 feet. also -1 isnt an elevation. If you were to draw a line youd put all the negative numbers on the bottom, and positive on the top.

7 0
3 years ago
Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3

T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)

Thus, T1(2) = 2 - 1

T1(2) = 1

T2 (2) = 2²/2 - 1/2

T2 (2) = 3/2

T2 (2) = 1.5

T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)

T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

f(2) = 1.386294

Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,

Since;

f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

Finally;

R3 <= |2/(4!)(2-1)^4|

R3 <= | 2 / 24× 1 |

R3 <= 1/12

R3 <= 0.083

5 0
3 years ago
3. Which sequence has a common difference of three? *
Andrej [43]

Answer:

2, 6, 18, 54..

Step-by-step explanation:

2, 6, 18, 54...

2 x 3 = 6

6 x 3 = 18

18 x 3 = 54

54 x 3 = ...

8 0
3 years ago
Read 2 more answers
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