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Damm [24]
3 years ago
5

The sum of the digits of a two-digit number is 11. The tens digit is one less

Mathematics
1 answer:
MakcuM [25]3 years ago
8 0
x+y=11\\
x+1=3y\ \ \ \ --->\ \ \ \ x=3y-1\\\\
3y-1+y=11\\
4y-1=11\ \ \ | add\ 1\\
4y=12\ \ \ | divide\ by\ 4\\
y=3\\\\
x=8\\\\
Original\ number \ is\ 83.
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Determine the quotient q(x) and remainder r(x) when f(x)=18x^4+27x^3+39x^2+22x+11
SVETLANKA909090 [29]

18x^4=3x^2\cdot6x^2, and

6x^2(3x^2+3x+4)=18x^4+18x^3+24x^2

Subtracting this from f(x) gives a remainder of

(18x^4+27x^3+39x^2+22x+11)-(18x^4+18x^3+24x^2)=9x^3+15x^2+22x+11

9x^3=3x^2\cdot3x, and

3x(3x^2+3x+4)=9x^3+9x^2+12x

Subtracting this from the previous remainder gives a new remainder of

(9x^3+15x^2+22x+11)-(9x^3+9x^2+12x)=6x^2+10x+11

6x^2=3x^2\cdot2, and

2(3x^2+3x+4)=6x^2+6x+8

Subtracting this from the previous remainder gives a new remainder of

(6x^2+10x+11)-(6x^2+6x+8)=4x+3

4x is not divisible by 3x^2, so we're done. We ended up with

\dfrac{f(x)}{g(x)}=\underbrace{6x^2+3x+2}_{q(x)}+\underbrace{\dfrac{4x+3}{3x^2+3x+4}}_{r(x)}

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I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be
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Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: g'(-1)\,(1) since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: g'(-1)= 7 as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

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Please help 15 points
Gwar [14]

Answer:

46/100

Step-by-step explanation:

The decimal reaches a point into the hundreths.

<em>kiniwih426</em>

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3 years ago
Read 2 more answers
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